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

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

- Trinomial
- Quadratic Formula
- Completing the Square

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:
- $Y=ax^2$
- $Y=ax^2+bx$
- $Y=ax^2+c$

Solve the following equation:

\( x^2+5x+4=0 \)

$Y=ax^2+bx+c$

Find $2$ numbers that satisfy the following two conditions:

$first~number \cdot second~number = c \cdot a$

$first~number + second~number = b$

**How do we proceed?**

First of all, let's write down on the side:

Then–

- We will find all the numbers whose product is $a\cdot c$ and write them down.
- We will check which pair of numbers from the pairs we found earlier will give us the sum $b$.
- We will write the pair of numbers that meets the $2$ conditions in this form:

$0= (x+one~solution)(x+second~solution)$

or with subtraction operations. - $2$ solutions of the quadratic equation will be those that solve the equation above (we will reverse the sign for the pair of numbers we found)

Tip – It is recommended to use the trinomial method when $a=1$

For more reading and practice on trinomials, click here!

**And now let's practice!**

Solve the quadratic function in front of you using trinomial:

$x^2-9x+14$

Solution:

First of all, let's write down on the side:

Let's find all the numbers whose product is $14$ (and remember the negative numbers as well)

We get:

$14,1$

$7,2$

$-14, -1$

$-7 ,-2$

Now, let's check which pair of numbers from the pairs we found earlier will give us the sum (-9)

The pair of numbers that managed to meet both conditions is $-7 ,-2$

We write the factorization:

$(x-7)(x-2)=0$

The solutions:

$X=7$

$X=2$

$Y=ax^2+bx+c$

Meet the quadratic formula:

$x = {-b \pm \sqrt{b^2-4ac} \over 2a}$

All you need to do is arrange the parameters of the quadratic function, substitute into the equation once with a plus and once with a minus, and find the solutions.

To learn more about the quadratic formula, click here!

**Let's practice!**

In front of us is the quadratic function:

$2x^2-3x+1$

Let's solve it using the quadratic formula:

First, let's arrange the parameters:

$a=2$

$b=-3$

$c=1$

Now, we substitute into the quadratic formula:

For the first time with plus-

$\frac{-(-3)+\sqrt{(-3)^2-4*2*1}}{2*2}=$

$=\frac{3+\sqrt1}4=\frac{4}4=1$

The second time with minus:

$\frac{-(-3)-\sqrt{(-3)^2-4*2*1}}{2*2}=$

$=\frac{3-\sqrt1}4=\frac{2}4=\frac{1}2$

We got $2$ solutions –

$X=\frac{1}2,1$

Test your knowledge

Question 1

Solve the following equation:

\( x^2+9x+8=0 \)

Question 2

Solve the following equation:

\( 2x^2-10x-12=0 \)

Question 3

Solve the equation

\( 3x^2-39x-90=0 \)

To use the method of completing the square, let's recall some of the formulas for the shortened multiplication:

$a^2+2ab+b^2=(a+b)^2$

$a^2-2ab+b^2=(a-b)^2$

**Solution method with example:**

Here is the function $4x^2+8x-5$

- Let's look at the quadratic function and focus only on $ax^2+bx$.

For now, we'll ignore $c$. In the example, we'll focus on $4x^2+8x$ - Let's recall the formulas for completing the square and ask, what expression can we put inside the parentheses squared, that is, which $(a-b)^2$ or $(a+b)^2$ as relevant, that will give us what appears in the pair we focused on $ax^2+bx$

In the example-

The appropriate formula for completing the square is:

$a^2+2ab+b^2=(a+b)^2$

Let's ask, what should we put as $a$ and $b$ to get $4x^2+8x$?

The answer is - $(2x+2)^2$

Let's expand this expression according to the formula for completing the square and get:

$4x^2+8x+4$

- Note that the expression in parentheses also brings with it a certain number and not just the pair we focused on, so we will need to cancel it.

If the added number is negative, we will add it to the equation to cancel it, and if the number is positive, we will subtract it from the equation to cancel it.

Additionally, we will return to $C$ from the original function and write it in the equation as well.

In the example:

$4x^2+8x+4$

The number $4$ was added. To cancel it, we will subtract $4$ and not forget the original $C$

$4x^2+8x-4-5=$ - We will substitute the pair $ax^2+bx$ with the appropriate expression in parentheses squared that we found and arrange the equation – we will get the completion of the square.

In the example:

$(2X+2)^2-4-5=$

$(2x+2)^2-9$

Now let's set the equation to 0 and solve:

$(2x+2)^2-9=0$

$(2x+2)^2=9$

$(2x+2)^2=3^2$

and also

$(2x+2)^2=(-3)^2$

Solution 1:

$2x+2=3$

$2x=1$

$x=\frac{1}2$

Solution 2:

$2x+2=-3$

$2x=-5$

$x=-2\frac{1}2$

For more information on completing the square, click here!

Do you know what the answer is?

Question 1

Solve the following equation:

\( -2x^2+22x-60=0 \)

Question 2

a = coefficient of x²

b = coefficient of x

c = coefficient of the independent number

Identifies a,b,c

\( 5x^2+6x-8=0 \)

Question 3

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?

a = coefficient of x²

b = coefficient of x

c = coefficient of the constant term

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

**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.**

$c=0$

Solve the following equation:

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

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

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

Solve the following equation:

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

Let's recall the quadratic formula:

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

$x_1=6$ $x_2=-1$

$x^2+9=0$

Solve the equation

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

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

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 (

$(\textcolor{red}{a}-\textcolor{blue}{b})^2=\textcolor{red}{a}^2-\underline{2\textcolor{red}{a}\textcolor{blue}{b}}+\textcolor{blue}{b}^2$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:

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

__So the correct answer is answer a.__

$y=8$

Related Subjects

- Quadratice Equations and Systems of Quadraric Equations
- Quadratic Equations System - Algebraic and Graphical Solution
- Solution of a system of equations - one of them is linear and the other quadratic
- Intersection between two parabolas
- Word Problems
- Properties of the roots of quadratic equations - Vieta's formulas
- Ways to represent a quadratic function
- Various Forms of the Quadratic Function
- Standard Form of the Quadratic Function
- Vertex form of the quadratic equation
- Factored form of the quadratic function
- The quadratic function
- Quadratic Inequality
- Parabola
- Symmetry in a parabola
- Plotting the Quadratic Function Using Parameters a, b and c
- Finding the Zeros of a Parabola
- Methods for Solving a Quadratic Function
- Completing the square in a quadratic equation
- Squared Trinomial
- The quadratic equation
- Families of Parabolas
- The functions y=x²
- Family of Parabolas y=x²+c: Vertical Shift
- Family of Parabolas y=(x-p)²
- Family of Parabolas y=(x-p)²+k (combination of horizontal and vertical shifts)