# number of solutions of a quadratic equation

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A quadratic expression (in Latin quadratus means squared) is an expression that has a squared term (a variable multiplied by itself) and is usually written in a standard form ax2+bx+c where a≠0 (x is a variable and a, b and c are constants).

Quadratic functions and equations questions may ask you to rewrite a quadratic function to one of its 3 forms or to solve a quadratic equation using a formula. Rewriting the quadratic function allows to display its different features: the coordinates of the vertex of the parabola and the x intercepts.

Finding the function of the parabola from its graph topic includes questions about finding the standard form, the quadratic form and the vertex form equations using the vertex coordinates and the x and y intercepts coordinates from the graph of the function.

Other questions include two types of word problems with quadratic functions:  calculating the area of a rectangle problems and calculating height versus time problems. The skills on this subject require writing a quadratic function from a word problem and solving a given quadratic function according to a word problem.

The formula sheet for quadratic equations is given below.

Quadratic equations and quadratic functions formula sheet is given below, it includes 3 types of basic formulas:

The forms of quadratic functions– standard form, factored form and vertex form.

Special factoring formulas– square of sum, square of difference and difference of squares.

Quadratic formula and determining the number of solutions– no solution, one solution or two solutions.

Note that all the formulas are explained in detail on this page. ### There are 3 forms for writing a quadratic function:

Standard form of a quadratic function:
y=ax2+bx+c

For example: y=x2-2x-8

Factored form of a quadratic function:
y=(ax+c)(bx+d)

For example: y=(x+2)(x-4)

Vertex form of a quadratic function:
y=a(x-h)2+k

For example: y=(x-1)2-9

The standard form is showing the y intercept of the graph, it is written as y=ax2+bx+c.
For example: y=x2-2x-8 is a standard form a quadratic function.

The factored form is showing the x intercepts of the graph, it is written as y=(ax+c)(bx+d). For example: The factored form of the function given above is y=(x+2)(x-4), therefore we know that the function intercepts the x axis at the points x=-2 and x=4 (y=0).
Note that you can open brackets of y=(x+2)(x-4) and get its standard form y=x2-2x-8.

The vertex form is showing the coordinates of the vertex of the parabola (the main point of a parabola is its vertex), it is written as y=a(x-h)2+k.
For example: The vertex form of the function given above is y=(x-1)2-9, therefore we know that the function has a vertex at the point x=1; y=-9.

Note that:
1. a≠0, otherwise the expression will be linear and not quadratic in a form of y=bx+c.
2. X is a variable and a, b, c, d, h and k are constants.
3. The letters a, b and c are used to represent constant values separately in every formula. Meaning that the constants a, b and c in the formula y=ax2+bx+c are different from the constants a, b and c in the formula y=(ax+c)(bx+d) and are different from the constant a in the formula y=a(x-h)2+k.

### Finding the factored form of a quadratic function

The types of questions on this subject can be:

1. We are given a quadratic function in a standard form as y=ax2+bx+c and we need to write it in a factored form as y=(ax+c)(bx+d).

2. We are given a quadratic function in a standard form as y=ax2+bx+c and we are asked to find the values of the x intercepts of the graph.

#### How can we find the factored form of a quadratic function with a=1?

In these questions we are given a simplified quadratic function y=1*x2+bx+c equal to y=x2+bx+c (a=1) and we are asked to find the factored form of this simplified function.

If we are  given the standard form y=x2+bx+c (the coefficient of x2 is 1), the factored form will be y=(x+c)(x+d) (the coefficients of x are also 1). This is why:

When we open brackets in the factored form, we multiply x by x to get x2 (this is the first expression it the FOIL formula- ax*bx). If the coefficients of the x variables in the factored form are different from 1 we can’t get 1 as a coefficient of x2 in the standard form. Therefore if we are given a structure of the standard form as y=x2+bx+c (meaning that a=1) then the structure of the factored form must be y=(x+c)(x+d) (meaning that a=1 and b=1).

For example: (x+2)(x-1)=x2-x+2x-2= x2+x-2. The coefficients of x in both parts of the factored form are 1 and the coefficient of x2 in the standard form is 1.

For example: (2x+2)(3x-1)=6x2-2x+6x-2= x2+4x-2. The coefficients of x in the factored form are a=2; b=3 and the coefficient of x2 in the standard form is a=6 (and not 1).

How to find the factored form given the standard form?

The factored form with FOIL formula is y=(x+c)(x+d)= x2+dx+cx+cd= x2+(c+d)x+cd.
The standard form is x2+bx+c.

We see that the constant c in the standard form is calculated from the multiplication of the constants c and d of the factored form. We also see that the constant b in the standard form is calculated from the sum of the constants c and d in the factored form. Therefore, if we are given the standard form, we know its b and c values and we can find the values of c and d of the factored form.

Consider this example:

Find the factored form of the function y=x2-2x-8.

In this example we have a standard form of y=x2-2x-8 (a=1, b=-2, c=-8).

Since a=1 in the standard form we know that a=1 and b=1 in the factored form. Therefore the factored form is y=(x+c)(x+d).

We know that c=-8 is equal to c*d in the factored form.

We know that b=-2 in the standard form is equal to c+d in the factored form.

What are the 2 numbers whose product is equal to -8 and whose sum is equal to -2? The numbers are -4 and 2. Therefore the factored form is (x+2)(x-4).

Checking by opening brackets with FOIL formula:
(x+2)(x-4)= x2-4x+2x-8= x2-2x-8. We see that the factored function is the same as the given standard function.

#### Finding the factored form of a quadratic function with a≠1: factoring out a common factor

We can use this method only if we can factor out a common factor so that we will be left with a function in which the coefficient of  x2 is 1.

Consider the following example:

Find the factored form of the function y=2x2-6x-8.

Note that the coefficient of x2 is not 1, it is 2. In this example we can factor out a common factor and stay with 1 as x2 coefficient:
y=2x2-6x-8=2(x2-3x-4)

Now we have a new standard form of y=x2-3x-4 (a=1, b=-3, c=-4).

Since a=1 in the new standard form we know that a=1 and b=1 in the factored form. Therefore the factored form is y=(x+c)(x+d).

We know that c=-4 is equal to c*d in the factored form.

We know that b=-3 in the standard form is equal to c+d in the factored form.

What are the 2 numbers whose product is equal to -4 and whose sum is equal to -3? The numbers are -4 and 1. Therefore the factored form is (x-4)(x+1).

Remember that the functions was y=2x2-6x-8=2(x2-3x-4) and we found the factored function for x2-3x-4. Therefore the factored function for the original function is 2(x-4)(x+1).

Check by opening brackets with FOIL formula:
2(x-4)(x+1)= 2(x2+x-4x-4)=2(x2-3x-4)=2x2-6x-8. We see that the factored function is the same as the given standard function.

#### Finding the factored form of a quadratic function with a≠1 without factoring out a common factor

These are the most difficult questions, since we can’t factor out a common factor and be left with a function in which the coefficient of  x2 is 1. Since the x2 coefficient is no longer 1 it is not enough to find 2 numbers whose sum equals b and whose multiplication equals c. We need to check the options for 2 numbers that also considering the x2 coefficient that is a.

If you are given the answers in multiple choice question you can check every answer by opening brackets with FOIL formula instead of solving the given function.

Consider the following example:

Find the factored form of the function y=4x2+25x+6.

Since the x2 coefficient is no longer 1 it is not enough to find 2 numbers whose sum is 6 and whose multiplication is 25. We need to check the options for 2 numbers that also considering the x2 coefficient that is 4.

We know that there are 2 options for the factored form so that x2 coefficient will be 4:
Option 1 is (4x±a )(x±b)   In this option multiplying the first numbers will get 4*1*x2=4x2.
In this option the coefficient of x is calculated by 4bx+ax, meaning x(4b+a) which is 4 times number b and number a.

Option 2 is (2x±a )(2x±b)  In this option multiplying the first numbers will get 2*2*x2=4x2.
In this option the coefficient of x is calculated by 2bx+2ax, meaning x(ab+2a) which is 2 times number b and 2 times number a.

The question asked should be: what are the numbers whose multiplication is 6 and one the following:
4 times number b and number a is 25: 4b+a=25
2 times number b and 2 times number a is 25: 2a+2b=25

The numbers whose multiplication is 6 are 1 and 6 or 2 and 3. For each one of these two options we also need to look at the second criteria and find a match.

If the numbers are 6 and 1 (6*1=6):
4 times number b and number a is 25: 4b+a=25

This is possible since 6*4+1=25, therefore the numbers are 6 and 1 and the coefficients of x2 are 6 and 4. The factored form is (4x+1)(x+6).

Checking the answer by opening brackets with FOIL formula: (4x+1)(x+6)=4x2+24x+x+6=4x2+25x+6.

Let’s check the other option: 2 times number b and 2 times number a is 25: 2a+2b=25
We can’t get this with the numbers 1 and 6 since 2*1+2*6=14 and not 25.

If the numbers are 2 and 3 (2*3=6):
4 times number b and number a is 25: 4b+a=25
We can’t get this with the numbers 2 and 3 since 2*1+3*4=16 and not 25, also 3*1+2*4=11 and not 25.

Let’s check the other option: 2 times number b and 2 times number a is 25: 2a+2b=25
We can’t get this with the numbers 2 and 3 since 2*2+3*2=10 and not 25.

#### How can we find the factored form of a quadratic function using special factoring?

In order to use special factoring, we need to remember the following formulas:

Square of sum formula:
(a+b)2=a2+b2+2ab

For example: (2x+5)2=4x2+25+20x

Square of difference formula:
(a-b)2=a2+b2-2ab

For example: (2x-5)2=4x2+25-20x

Difference of squares formula:
a2-b2=(a+b)(a-b)

For example: 4x2-25=(2x+5)(2x-5)

In this solving method we will be given the result of one of these 3 formulas so we can factor it immediately using one of the formulas. In order to do it we need to look for perfect squares.
Perfect squares are integers which square is an integer. Note that the numbers a2 and b2 are perfect squares since their squares a and b are also integers.

Steps for finding the factored form of a quadratic function using special factoring:
1. Factor out common factors if it will help you to get perfect squares for the next step.
2. Find the components of the functions a2+b2±2ab or a2-b2 and do special factoring using the formulas:

When we see a sum of 2 numbers that are perfect squares, we know that they can be a2 and b2 and we need to see if we also have the third part of the formula which is 2ab or -2ab to complete the formula a2+b2+2ab=(a+b)2 or a2+b2-2ab =(a-b)2.

When we see a difference of 2 numbers that are perfect squares, we know that they can be a2 and b2 and we factor the formula a2-b2=(a+b)(a-b).

Consider the following example:

What is the factored form of a function f(x)=8x4+2x2+8x3?

Step 1: Factor out common factors:
We can factor out 2 as a common factor: 8x4+2x2+8x3=2(4x4+x2+4x3).

Step 2: We see that 4x4 is a perfect square since its square root is an integer: √4x4=2x2.
We see that x2 is a perfect square since its square root is an integer: √x2=x.
We see that the multiplication of the 2 perfect squares by 2 gives us the third number 2x2*x*2=4x3.

One of the numbers is negative so we know that we can do a special factoring using the formula a2-b2=(a+b)(a-b) getting  2(4x4+x2+4x3)=2(2x2+x)2.

Note that we can’t solve without factoring out 2 as a common factor since 8x4 and 2x2 are not perfect squares. In this case we must factor out a common factor before the perfect factoring.

Consider the following example:

What is the factored form of a function f(x)=16x4-4x2?

Step 1: Factor out common factors:
We can factor out 4 as a common factor: 16x4-4x2=4(4x4-x2).

Step 2: We see that 4x4 is a perfect square since its square root is an integer: √4x4=2x2.
We see that x2 is a perfect square since its square root is an integer: √x2=x.

One of the numbers is negative and so we know that we can do a special factoring using the formula a2-b2=(a+b)(a-b) getting  4(4x2-x2)=4(2x2+x)(2x2-x).

Note that factoring out 2 as a common factor will get numbers 8x4 and 2x2 that are not perfect squares 16x4-4x2=2(8x4-2x2), therefore it is not suitable because it makes the next step impossible.

Note that we can solve without factoring out 4 as a common factor since 16x4 and 4x2 are also perfect squares. In this case we will need to factor out a common factor after the perfect factoring.
16x4-4x2=(4x2+2x)(4x2-2x)=2(2x2+x)*2(2x2-x)=2*2*(2x2+x)(2x2-x)=4(2x2+x)(2x2-x).

Consider the following example:

What is the factored form of a function f(x)=20x2-5?

Step 1: Factor out common factors:
We can factor out 5 as a common factor: 20x2-5=5(4x2-1).

Step 2: We see that 4x2 is a perfect square since its square root is an integer: √4x2=2x.
We see that 1 is a perfect square since its square root is an integer: √1=1.

One of the numbers is negative and so we know that we can do a special factoring using the formula a2-b2=(a+b)(a-b) getting  5(4x2-1)=5(2x+1)(2x-1).

Note that we can’t solve without factoring out 5 as a common factor since 20x2 and 5 are not perfect squares. In this case we must factor out a common factor before the perfect factoring.

#### How can we find the x intercepts of a quadratic function using its factored form?

We know that the y values of the interception points with the x axis are y=0. Therefore we know that at the factored form equation at the x axis interception points is (x+c)(x+d)=0.

The zero product property states that if the multiplication of 2 numbers is equal to 0 then one of the numbers must be equal to 0. In the factored formula equation that means that x+c= 0 or x+d=0. Since c and d are constants that we know we can see the x values from the factored formula. For example: in the function y=(x+2)(x-4)=0, the intercepts with the x axis are at the points x=-2 and x=4.

Consider the following example:

Find the x intercepts of the function f(x)=x2+3x-10.

Step 1: we need to find the factored form of the function. Since the standard form the is simplified to y=x2+bx+c (a=1) we know that the factored form will be (x+c)(x+d) (so that a=1, b=1).
The coefficient -10 in the standard formula is calculated from the multiplication of c and d in the factored formula.
The coefficient 3 in the standard formula is calculated from the sum of c and d in the factored formula.

What are the 2 numbers whose product is equal to -10 and whose sum is equal to 3? The numbers are -2 and 5. Therefore the factored form is (x-2)(x+5).

Check by opening brackets with FOIL formula:
(x-2)(x+5)= x2+5x-2x-10= x2-3x-10. We see that the factored function is the same as the given standard function.

Step2: We need to solve the equation (x-2)(x+5)=0.
According to the zero product property we know that the function has 2 x intercepts x=2 and x=-5.

The following graph presents the graph of the function with its intercepts. ### Finding the vertex form of a quadratic function

The types of questions on this subject can be:
1. We are given a quadratic function in a standard form as y=ax2+bx+c and we need to write it in a vertex form as y=a(x-h)2+k.
2. We are given a quadratic function in a standard form as y=ax2+bx+c and we are asked to find the values of the vertex of the graph.

How can we find the vertex form of a quadratic function?
If we look at the vertex formula y=a(x-h)2+k we see the quadratic expression (x-h)2.
(x-h)2=(x-h)(x-h)

Opening brackets with FOIL formula will give us:
(x-h)(x-h)=x2-hx-hx+h2= x2-2hx+h2 (h is constant).

Note that the form of x2-2hx+h2 is structured like the standard form y=ax2+bx+c. There is a squared variable, a constant and a multiplication of a constant and a variable.

How can we find a vertex form from a standard form? We will rewrite the standard form to fit to the vertex form.

#### Finding the vertex form of a function with a minus sign in the x coefficient

Consider the following example:

The standard form is y= x22x-8. What is the vertex form of the function?

Look at the expression x2-2x-8. What do we need to do in order to write it in a form of a(x-h)2= x2-2hx+h2?

Making x2-2x-8 to be x2-2hx+h2:
We have x2 as x2
We have -2x as -2hx
We have -8 as h2

Therefore, we see that if h=1 well have an equality in x2 and -2x and what is left is only h2. his supposed to be 1 and it is -8.

That means we have x2-2x+1 and we need minus 9 to finish the standard form:
x2-2x-8=x2-2x+1-9

Now we can write x2-2x+1 as a square equation:
x2-2x+1=(x-1)2

Writing the standard form again will give us the vertex form:
x2-2x-8=x2-2x+1-9=(x-1)2-9

The vertex formula is y=a(x-h)2+k therefore the constants are a=1, h=1 and k=-9.

Checking the answer: we can open brackets of the vertex form to see that we get the standard form:
(x-1)2-9=(x-1)(x-1)-9=x2-x-x+1-9= x2-2x-8.

#### Finding the vertex form of a function with a plus sign in the x coefficient

Consider the following example:

The standard form is y=x2+2x+10. What is the vertex form of the function?

Look at the expression x2+2x+10, since we have a plus sign before 2x we will write the vertex form as a(x+h)2=x2+2hx+h2 and deal with the sign later.

Making x2+2x+10 to be x2+2hx+h2:
We have x2 as x2
We have 2x as 2hx
We have 10 as h2

Therefore, we see that if h=1 well have an equality in x2 and 2x and what is left is only h2. his supposed to be 1 and it is 10.

That means we have x2+2x+1 and we need 9 to finish the standard form:
x2+2x+10=x2+2x+1+9

Now we can write x2+2x+1 as a square equation:
x2+2x+1=(x+1)2

Writing the standard form again will give us the vertex form:
x2+2x+10=x2+2x+1+9=(x+1)2+9
The vertex formula is y=a(x-h)2+k therefore the constants are a=1, h=-1 and k=9.

Note that in order to get x+1 from the formula x-h we need h to be h=-1 and not h=1, meaning that x-h=x – -1=x+1.

Checking the answer: we can open brackets of the vertex form to see that we get the standard form:
(x+1)2+9=(x+1)(x+1)+9=x2+x+x+1+9= x2+2x+10.

#### How can we find the vertex values of a quadratic function using its vertex form?

The vertex form y=a(x-h)2+k has 2 expressions a(x-h)2 and k. Note that the expression in the brackets has a quadratic exponent therefore it is positive for every x.

The x coordinate of the vertex is equal to h and the y coordinate of the vertex is equal to k. This is why:

If a is positive then the function has a minimum point. At the minimum point, the y value of the function must be the smallest. If the first expression a(x-h)is not equal to 0, it will cause the y value to be bigger and we want it to be the smallest, therefore at the minimum point the first expression a(x-h)2 must be equal to 0 so that what is left is only the constant k. What are the conditions for a(x-h)2=0? We need x to be equal to h and then y will be equal to k.

If a is negative then the function has a maximum point. At the maximum point, the y value of the function must be the biggest. Note that since a is negative the first expression a(x-h) 2  will be always negative. If the first expression a(x-h)2 is not equal to 0, it will cause the y value to be smaller and we want it to be the biggest, therefore at the maximum point the first expression a(x-h)2 must be equal to 0 so that what is left is only the constant k. What are the conditions for a(x-h)2=0? We need x to be equal to h and then y will be equal to k.

Consider the following example:

What are the coordinates of the vertex in the function (x-1)2-9?

Since the coefficient of x2 is positive, we know that the function has a minimum point.
In order for y to be the smallest possible we need the expression (x-1)2  to be 0.
This is possible when x=1. Therefore, at the vertex point x=h=1 and y=k=-9.

The following graph presents the graph of the function with its vertex point. Consider the following example:

What are the coordinates of the vertex in the function (x+1)2+9?

Since the coefficient of x2 is positive, we know that the function has a minimum point. In order for y to be the smallest possible we need the expression (x+1)2  to be 0.
This is possible when x=-1.
Therefore, at the vertex point x=h=-1 and y=k=9.

The following graph presents the graph of the function with its vertex point. A quadratic equation is an equation that has a squared term (a variable multiplied by itself) and is usually written in a standard form ax2+bx+c=0 where a≠0 (x is a variable and a, b and c are constants). For example:

3x2+7x-1=0 is a quadratic equation with the parameters a=3, b=7 and c=-1.

-x2+7x=0 is a quadratic equation with the parameters a=-1, b=7 and c=0.

-5x2-1=0 is a quadratic equation with the parameters a=-5, b=0 and c=-1.

7x-1=0 is a not a quadratic equation since a=0, this equation is linear.

#### Solving a quadratic equation without x term (b=0)

Since we have only 2 numbers in the equation, we can isolate x2 and solve by taking a square root from each side of the equation.

Note that when we take square root, we get 2 answers, a positive answer and a negative answer. This is because a square of a negative number is positive. Therefore, don’t forget the negative answer! For example: √4x2=2x or -2x, since (2x)2=4x and also (-2x)2=4x.

Steps for solving a quadratic equation without x term (b=0):
1. Isolating x.
2. Taking a square root from each side of the equation.
3. Don’t forget that there is a positive answer and a negative answer.

Consider this example:

Solve 4x2-16=0.

4x2-16=0
4x2=16
4x2/4=16/4
x2=4
√x2=√4
x=2, x=-2

Note that x=-2 is also a solution for the equation.

#### Solving a quadratic equation without the constant term (c=0)

In these questions we need to solve a quadratic equation in its standard form that is ax2+bx (c=0). We can factor out x as a common factor getting x(ax+b)=0.
The zero product property states that if the multiplication of 2 numbers is equal to 0 then one of the numbers must be equal to 0, therefore x=0 or ax+b=0. We can solve ax+b=0 since it is a simple linear equation.

Consider the following example:

What are the solutions for the equation 2x2-4x=0?

2x2-4x=0
x(2x-4)=0
x=0 or 2x-4=0
2x=4
x=2

For x=0 solution: 2x2-4x=2*02-4*0=0
For x=2 solution: 2x2-4x=2*22-4*2=2*4-8=8-8=0

#### Solving a quadratic equation by factoring

We need to solve a quadratic equation in its standard form that is ax2+bx+c=0. Instead we can write it in its factored form (ax+c)(bx+d)=0.

The zero product property states that if the multiplication of 2 numbers is equal to 0 then one of the numbers must be equal to 0. In the factored formula equation (ax+c)(bx+d)=0 that means that ax+c= 0 or bx+d=0. Since a,b c and d are constants that we know we can find the x values from the factored formula by solving this 2 linear equations. For example: in the function y=(x+2)(x-4)=0, the intercepts with the x axis are at the points x=-2 and x=4.

Steps for solving a quadratic equation with factoring:
Step 1: Rewrite the standard form of the quadratic equation ax2+bx+c=0 as its factored form (ax+c)(bx+d)=0.
Step 2: Solve 2 linear equations by making each equation equal to 0: ax+c= 0 or bx+d=0. The solutions to the factored form equation are also the solutions to the standard form equation.

Consider the following example:

What are the solutions of the equation 2x2-6x-8=0?

Step 1: Rewriting the standard form of the quadratic equation to its factored form.
This was explained before.
First we can factor out 2 as a common factor getting 2(x2-3x-4)=0
We know that the factored form is (x+a)(x+b).
We know that ab=-4
We know that bx+ax=-3x, x(b+a)=-3x, b+a=-3.

The numbers which multiplication is equal to -4 and their sum is equal to -3 are -4 and 1, therefore the factored form is 2(x-4)(x+1).

Step 2: Solving the equations:
We will write the factored form as equal to zero 2(x-4)(x+1)=0. According to the zero product property we know that x-4=0 or x+1=0. Meaning that x=4 or x=-1 are the solutions for the equation.

2(4-4)(x+1)=2*0*(x-1)=0
2(x-4)(-1+1)=2*(x-4)*0=0

x=(-b±√(b2-4ac))/2a

For example: x2+x-30=0
a=1, b=1, c=-30

x=(-1±√(12-4*1*-30))/(2*1))
x1=(-1+11)/2=5
x2=(-1-11)/2=-6

For a standard form of a quadratic equation ax2+bx+c=0 the quadratic formula is
x=(-b±√(b2-4ac))/2a

Step 1: Write the quadratic equation in its standard form ax2+bx+c=0.
Step 2: Plug the constants a, b and c into the quadratic formula x=(-b±√(b2-4ac))/2a.
Step 3: Evaluate x by solving the formula. Note that since we have a square root in the formula, we will have 2 solutions √(b2-4ac) and -√(b2-4ac).

Consider the following example:

What are the solutions to the equation x2+x=30?

Step 1: Writing the quadratic equation in its standard form
x2+x=30
x2+x-30=0
a=1, b=1, c=-30

Step 2: Plugging the constants a, b and c into the quadratic formula
x=(-b±√(b2-4ac))/2a
x=(-1±√(12-4*1*-30))/(2*1))
x=(-1+√(1+4*1*30))/2

Solution 1: (-1+√121)/2=(-1+11)/2=10/2=5
Solution 2: (-1-√121)/2=(-1-11)/2=-12/2=-6

x2+x-30=0
52+5-30=0, 0=0
(-6)2-6-30=0, 36-6-30=0, 0=0

### Determining the number of the solutions of a quadratic equation

No real solution:
If b2-4ac<0 the quadratic equation has no real solution

For example: a=2, b=1, c=1
b2-4ac=1-4*2*1=-7<0 (no solution)
x=(-b±√-7)/2a (no solution)

1 real solution:
If b2-4ac=0 the quadratic equation has 1 real solution:
x=-b/2a

For example: a=1, b=2, c=1
b2-4ac=4-4*1*1=0 (one solution)
x=(-b±√(0)/2a
x=(-2±√0)/2=-1

2 real solutions:
If b2-4ac>0 the quadratic equation has 2 real solutions:
x1=(-b+√(b2-4ac))/2a
x2=(-b-√(b2-4ac))/2a

For example: a=1, b=-2, c=-8
b2-4ac=4-*1*-8=36>0 (two solutions)
x1=(-b+√(b2-4ac)/2a=(2+√36)/2=4
x2=(-b-√(b2-4ac)/2a=(2-√36)/2=-2

The quadratic equation can have 2 solutions, one solution or no solution. The number of the solutions depends on the value of the expression under the root b2-4ac. This expression is called discriminant

We know thar for a standard form of a quadratic equation ax2+bx+c=0 the quadratic formula is x=(-b±√(b2-4ac))/2a. When we evaluate x using this quadratic formula, we take a square root from the expression b2-4ac.

If the discriminant b2-4ac is negative we can’t take the root, because squared number is always positive. In this case the equation has no real solution.

If the discriminant b2-4ac is zero, the equation has one real solution that is x=-b/2a.

If the discriminant b2-4ac is positive, the equation has two real solutions x=(-b+√(b2-4ac))/2a and x=(-b√(b2-4ac))/2a.

Consider the following example:

What are the solutions of the equation x2=-2x-1?

Step 1: Writing the quadratic equation in its standard form:
x2=-2x-1
x2+2x+1=0
a=1, b=2, c=1

Step 2: Plugging the constants a, b and c into the quadratic formula:
x=(-b±√(b2-4ac))/2a
x=(-2±√(22-4*1*1))/(2*1)
x=(-2±√(4-4))/2
x=(-2±√0)/2
The discriminant is zero therefore we have 1 solution
x=-2/2=-1 the equation has only one solution x=-1.

x2=-2x-1
(-1)2=-2*-1-1
1=2-1
1=1

Consider the following example:

What are the solutions of the equation 2x2=-x-1?

Step 1: Writing the quadratic equation in its standard form:
2x2=-x-1
2x2+x+1=0
a=2, b=1, c=1

Step 2: Plugging the constants a, b and c into the quadratic formula:
x=(-b±√(b2-4ac))/2a
x=(-1±√(12-4*2*1))/(2*2)
x=(-1±√(1-8))/4

The discriminant is negative therefore the equatuon has no solution.
We can’t take a square root from -7 therefore the equation has no solution.

Consider the following example:

What are the solutions of the equation x2=2x+8?

Step 1: Writing the quadratic equation in its standard form:
x2=2x+8
x2-2x-8=0
a=1, b=-2, c=-8

Step 2: Plugging the constants a, b and c into the quadratic formula:
x=(-b±√(b2-4ac))/2a
x=(–2±√((-2)2-4*1*-8))/(2*1)
x=(2±√(4+32))/2
The discriminant is positive therefore we have 2 solutions
x=(2±√36)/2
x=(2±6)/2
We have two answers x1=8/2=4 or x2=-4/2=-2

x2-2x-8= 42-2*4-8=16-8-8=0

x2-2x-8= (-2)2-2*-2-8=4+4-8=0

#### Writing the factored form of a quadratic equation using the solutions from the quadratic formula

In order to write the factored form of the quadratic equation with the solutions we found using the quadratic formula we need to change the signs inside the brackets. If the solution is x=a in order for the expression (x+a) to be zero we need to write (x-a), that will give us a-a=0.

Consider the following example:

The equation x2-2x-8=0 has 2 solutions x=4 and x=-2. What is the factored form of the equation?

The factored form is (x-4)(x+2)=0, because plugging the solutions x=4 or x=-2 will give us zero as we need: (4-4)=0 or (2-2)=0.

We can also check the factored form by solving (x-4)(x+2).
(x-4)(x+2)=x2+2x-4x-8=x2-2x-8 we see that factored form is equal to the standard form.
The solutions for x2-2x-8=0 are x=4 or x=-2.
The solutions for (x-4)(x+2)=0 are x=4 or x=-2.

There are 2 main types of word problems with exponential functions:  calculating the area of a rectangle problems and calculating height versus time problems.

The skills required are:
Writing a quadratic function from a word problem.
Solving a given quadratic function according to a word problem.

#### Word problems- calculating the area of a rectangle

The formula of the area of a rectangle A=lw is its length l multiplied by its width w. If we receive data about the parameter l or w, we can express one parameter in terms of the other. Then we will have only one variable x and we will be able to write a factored form of a quadratic equation.

Consider the following example:

Rectangles length is 5 inches bigger than its width. What is the function representing its area?

w=x
l=x+5
A=x*(x+5)= x2+5x

#### Word problems- calculating height versus time

The relationship between height (h) and time (t) of falling objects is a quadratic function in its standard form h(t)=-at2+bt+c.
h(t)- the output- height (usually in feet)
t- the variable- time after the object is thrown (usually in seconds)
a coefficient that represents the object’s acceleration due to gravity (therefore a is negative)
b coefficient represents the initial velocity of the object
c- initial height of the object above the ground level (the height 0 is the ground, usually in feet)

Consider the following example:

The ball is launched up in the air 10 feet above the ground.
The function that models the height of the ball h, in feet, t seconds after being launched is h(t)=-t2+3t+10.

What is the height of the ball 0.05 minutes after it was launched?

We know that the time in the question is asked in minutes and the time in the formula is measured in seconds. 1 minute= 60 seconds, 0.05 minutes= 60*5/100 seconds= 3 seconds, therefore the time for the formula is t=3.
h(t)=-t2+3t+10
h(t=3)=-32+3*3+10=-9+9+10=10 feet.

Consider the following example:

The ball is launched up in the air 10 feet above the ground. The function that models the height of the ball h, in feet, t seconds after being launched is h(t)=-t2+3t+10. When will the ball be at the launching height?

The ball will go up and then fall down, the question is when will it be at the initial launching height? The launching height is 10, therefore we know that h(t)=-t2+3t+10=10.
-t2+3t+10=10
-t2+3t=0
t(-t+3)=0
t=0 or t=3

The answer t=0 is the launching point that we already know. The second answer t=3 is the answer. The ball will be again at the launching height of 10 feet after 3 seconds.

Consider the following example:

The ball is launched up in the air 10 feet above the ground. The function that models the height of the ball h, in feet, t seconds after being launched is h(t)=-t2+3t+10. How many seconds after launch will the ball hit the ground?

Since we are asked about the moment the ball will hit the ground, we know that the height h should be equal to 0.
h(t)=-t2+3t+10=0

We can solve with the quadratic formula x=(-b±√(b2-4ac))/2a
a=-1, b=3, c=10
x=(-3±√(32-4*-1*10))/(2*-1)
x=(-3±√(9+40))/-2
x=(-3±7)/-2
x=-10/-2=5 or x=4/-2=-2

Since we are looking for time the answer can’t be negative, therefore the only answer is t=5 seconds.

For x=5: -t2+3t+10=-52+3*5+10=-25+15+10=0

### Finding the function of the parabola from its graph

We can find the standard form, the quadratic form and the vertex form equations using the vertex coordinates and the x and y intercepts coordinates from the graph.

#### Finding the factored form of the quadratic function from its graph

Recall that the factored form of the quadratic function is y=a(x-b)(x-c), it shows the x intercepts of the parabola so that the intercepts are (b,0) and (c,0).

Given the x intercepts coordinates from the graph, we know the values of b and c and we can write them in the function y=a(x-b)(x-c).  We are left with 3 variables: x, y (we need them for the function equation) and a. In order to find a we can plug any point into the equation, except the y intercepts values that we used.

Consider the following example:

Find the factored form of the graph below. Step 1: Identifying the vertex and the x intercept points:
The vertex is (-1,-8)
The x intercepts are (1,0) and (-3,0)

Step 2: Writing b and c values in the function y=a(x-b)(x-c):
In y=a(x-b)(x-c) we have b=1 and c=-3 so that the function is y=a(x-1)(x – – 3)=a(x-1)(x+3)
Checking: if x=1 we have y=a*0*4=0
Checking: if x=-3 we have y=a*-4*0=0

Step 3: Finding the value of a:
We know that the function is y=a(x-1)(x+3) and what is left is to find the value of a.
We can plug x and y values of any point from the graph.
We can plug the vertex coordinates (-1,-8) and get
-8=a(-1-1)(-1+3)
-8=-2*2*a
-8=-4a
a=2

The factored function is y=2(x-1)(x+3).

#### Finding the vertex form of the quadratic function from its graph

Given the vertex coordinates from the graph, we know the values of h and k and we can write them in the function y=a(x-h)2+k. We are left with 3 variables x, y (we need them for the function equation) and a. In order to find a we can plug any point into the equation, except the vertex values that we used.

Consider the following example:

Find the vertex form of the graph below. Step 1: Identifying the vertex and the x intercept points:
The vertex is (-1,-8)
The x intercepts are (1,0) and (-3,0)

Step 2: Writing h and k values in the function y=a(x-h)2+k:
y=a(x-h)2+k
Since the vertex is (-1,-8) we know that k=-8 and h=-1.
y=a(x+1)2-8 (if we plug here x=-1 we get y=-8 and this is what we need for the vertex)

Step 3: Finding the value of a:
We know that the function is y=a(x+1)2-8 and what is left is to find the value of a.
We can plug x and y values of any point from the graph.
We can plug one of the x intercepts coordinates (1,0) and get
y=a(x+1)2-8
0= 4a-8
a=2

We can plug the other x intercept coordinates (-3,0) and get the same answer
y=a(x+1)2-8
0=4a-8
a=2

The vertex function is y=2(x+1)2-8

#### Finding the standard form of the quadratic function from its graph

We know the coordinates of 3 points from the graph: the vertex and the x intercepts. The standard form of a quadratic equation is y=ax2+bx+c. We can plug the (x,y) coordinates of the 3 point given in the graph and get 3 equations with 3 variables a, b and c. We can solve the equations and calculate the values of a, b and c.

Note that we can also find the standard form by finding the vertex form or the factored form and opening brackets.

Consider the following example:

Find the standard form of the graph below. Step 1: Identifying the vertex and the x intercept points:
The vertex is (-1,-8)
The x intercepts are (1,0) and (-3,0)

Step 2: writing 3 equations with 3 variables a, b and c:
y=ax2+bx+c
For (-1,-8) we get -8=a-b+c
For (1,0) we get 0=a+b+c
For (-3,0) we get 0=9a-3b+c

Step 3: solving the equations:
Isolating a from equation 1:
a=-8+b-c

Plugging a=-8+b-c into equation 2:
0=a+b+c
0=-8+b-c+b+c
2b=8
b=4
now we know that
a=-8+b-c=-8+4-c=-4-c

Plugging a=-4-c and b=4 into equation 3:
0=9a-3b+c
0=9(-4-c)-3*4+c
0=-36-9c-12+c
0=-48-8c
-8c=48
c=-6

Finding a by plugging b=4 and c=-6 into a=-8+b-c:
a=-8+4+6=2

The standard form is y=2x2+4x-6

#### Finding the standard form from the vertex form by opening brackets:

As we saw earlier the vertex form of the graph was is y=2(x+1)2-8

Opening brackets will get us
y=2(x+1)2-8
y=2(x2+2x+1)-8
y=2x2+4x+2-8

The standard form is y=2x2+4x-6

#### Finding the standard form from the factored form by opening brackets:

As we saw earlier the factored form of the graph was is y=2(x-1)(x+3)

Opening brackets will get us
y=2(x-1)(x+3)
y=2(x2+3x-x-3)
y=2(x2+2x-3)
y=2x2+4x-6 You just finished studying quadratic equations and quadratic functions topic, the first topic of passport to advanced mathematics subscore!

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