## Methods for solving a quadratic function

In this article, we will learn the three most common ways to solve a quadratic function easily and quickly.

1. Trinomial
3. Completing the Square

#### Reminder:

The basic quadratic function equation is:
$Y=ax^2+bx+c$

When:
$a$ - the coefficient of $X^2$
$b$ - the coefficient of $X$
$c$ - the constant term

• $a$ must be different from $0$
• $b$ or $c$ can be $0$
• $a,b,c$ can be negative/positive
• The quadratic function can also look like this:
• $Y=ax^2$
• $Y=ax^2+bx$
• $Y=ax^2+c$

## Examples with solutions for Solving Quadratic Equations

### Exercise #1

a = coefficient of x²

b = coefficient of x

c = coefficient of the constant term

What is the value of $c$ in the function $y=-x^2+25x$?

### Step-by-Step Solution

Let's recall the general form of the quadratic function:

$y=ax^2+bx+c$ The function given in the problem is:

$y=-x^2+25x$$c$is the free term (meaning the coefficient of the term with power 0),

In the function in the problem there is no free term,

Therefore, we can identify that:

$c=0$Therefore, the correct answer is answer A.

$c=0$

### Exercise #2

Solve the following equation:

$x^2+5x+4=0$

### Step-by-Step Solution

The parameters are expressed in the quadratic equation as follows:

aX2+bX+c=0

We substitute into the formula:

-5±√(5²-4*1*4)
2

-5±√(25-16)
2

-5±√9
2

-5±3
2

The symbol ± means that we have to solve this part twice, once with a plus and a second time with a minus,

This is how we later get two results.

-5-3 = -8
-8/2 = -4

-5+3 = -2
-2/2 = -1

And thus we find out that X = -1, -4

$x_1=-1$ $x_2=-4$

### Exercise #3

Solve the following equation:

$2x^2-10x-12=0$

### Step-by-Step Solution

We'll substitute the given data into the formula:

$x={{-(-10)\pm\sqrt{-10^2-4\cdot2\cdot(-12)}\over 2\cdot2}}$

Let's simplify and solve the part under the square root:

$x={{10\pm\sqrt{100+96}\over 4}}$

$x={{10\pm\sqrt{196}\over 4}}$

$x={{10\pm14\over 4}}$

Now we'll solve using both methods, once with the addition sign and once with the subtraction sign:

$x={{10+14\over 4}} = {24\over4}=6$

$x={{10-14\over 4}} = {-4\over4}=-1$

We've arrived at the solution: X=6,-1

$x_1=6$ $x_2=-1$

### Exercise #4

$x^2+9=0$

Solve the equation

### Step-by-Step Solution

The parameters are expressed in the quadratic equation as follows:

aX2+bX+c=0

We identify that we have:
a=1
b=0
c=9

We recall the root formula:

We replace according to the formula:

-0 ± √(0²-4*1*9)

2

We will focus on the part inside the square root (also called delta)

√(0-4*1*9)

√(0-36)

√-36

It is not possible to take the square root of a negative number.

And so the question has no solution.

No solution

### Exercise #5

$60-16y+y^2=-4$

### Step-by-Step Solution

Let's solve the given equation:

$60-16y+y^2=-4$First, let's arrange the equation by moving terms:

$60-16y+y^2=-4 \\ 60-16y+y^2+4=0 \\ y^2-16y+64=0$Now, let's note that we can break down the expression on the left side using the short quadratic factoring formula:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-2\textcolor{red}{a}\textcolor{blue}{b}+\textcolor{blue}{b}^2$This is done using the fact that:

$64=8^2$So let's present the outer term on the right as a square:

$y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-16y+\textcolor{blue}{8}^2=0$Now let's examine again the short factoring formula we mentioned earlier:

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$And the expression on the left side of the equation we got in the last step:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2=0$Let's note that the terms $\textcolor{red}{y}^2,\hspace{6pt}\textcolor{blue}{8}^2$indeed match the form of the first and third terms in the short multiplication formula (which are highlighted in red and blue),

But in order for us to break down the relevant expression (which is on the left side of the equation) using the short formula we mentioned, the match to the short formula must also apply to the remaining term, meaning the middle term in the expression (underlined):

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$In other words - we'll ask if it's possible to present the expression on the left side of the equation as:

$\textcolor{red}{y}^2-\underline{16y}+\textcolor{blue}{8}^2 =0 \\ \updownarrow\text{?}\\ \textcolor{red}{y}^2-\underline{2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}}+\textcolor{blue}{8}^2 =0$And indeed it holds that:

$2\cdot y\cdot8=16y$So we can present the expression on the left side of the given equation as a difference of two squares:

$\textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0$From here we can take out square roots for the two sides of the equation (remember that there are two possibilities - positive and negative when taking out square roots), we'll solve it easily by isolating the variable on one side:

$(y-8)^2=0\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ y-8=\pm0\\ y-8=0\\ \boxed{y=8}$

Let's summarize then the solution of the equation:

$60-16y+y^2=-4 \\ y^2-16y+64=0 \\ \downarrow\\ \textcolor{red}{y}^2-2\cdot\textcolor{red}{y}\cdot\textcolor{blue}{8}+\textcolor{blue}{8}^2=0 \\ \downarrow\\ (\textcolor{red}{y}-\textcolor{blue}{8})^2=0 \\ \downarrow\\ y-8=0\\ \downarrow\\ \boxed{y=8}$

$y=8$

### Exercise #6

a = Coefficient of x²

b = Coefficient of x

c = Coefficient of the independent number

what is the value of $a$ in the equation

$y=3x-10+5x^2$

### Video Solution

$a=5$

### Exercise #7

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

$x^2+4x-5=0$

What are the components of the equation?

### Video Solution

$a=1$ $b=4$ $c=-5$

### Exercise #8

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

$10x^2+5+20x=0$

What are the components of the equation?

### Video Solution

$a=10$ $b=20$ $c=5$

### Exercise #9

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

what is the value of $a$ in the equation

$y=-x^2-3x+1$

### Video Solution

$a=-1$

### Exercise #10

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

$5-6x^2+12x=0$

What are the components of the equation?

### Video Solution

$a=-6$ $b=12$ $c=5$

### Exercise #11

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

Identifies a,b,c

$5x^2+6x-8=0$

### Video Solution

$a=5$ $b=6$ $c=-8$

### Exercise #12

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

$-8x^2-5x+9=0$

What are the components of the equation?

### Video Solution

$a=-8$ $b=-5$ $c=9$

### Exercise #13

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

$-x^2-2=0$

What are the components of the equation?

### Video Solution

$a=-1$ $b=0$ $c=-2$

### Exercise #14

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

$x^2+7x=0$

What are the components of the equation?

### Video Solution

$a=1$ $b=7$ $c=0$

### Exercise #15

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

what is the value of $b$in this quadratic equation:

$y=4x^2-16$

### Video Solution

$b=0$