Methods for solving a quadratic function

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

Test yourself on solution of quadratic equation!

What is the value of X in the following equation?

$$X^2+10X+9=0$$

How do you solve a quadratic function using trinomial?

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

*Relevant notation in the Word file*

Then–

1. We will find all the numbers whose product is $a\cdot c$ and write them down.
2. We will check which pair of numbers from the pairs we found earlier will give us the sum $b$.
3. 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.
4. $2$ 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 $a=1$

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:

*Relevant notation in the Word file*

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)

*Relevant notation in the Word file*

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

We write the factorization:
$(x-7)(x-2)=0$
The solutions:
$X=7$
$X=2$

$Y=ax^2+bx+c$

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

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.

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 –
$X=\frac{1}2,1$

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How do you solve a quadratic function by completing the square?

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 $4x^2+8x-5$

1. 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 $4x^2+8x$
2. 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:
$4x^2+8x+4$

1. 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 $C$
$4x^2+8x-4-5=$
2. 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:
$2x+2=-3$
$2x=-5$
$x=-2\frac{1}2$