Methods for solving a quadratic function

🏆Practice solution of quadratic equation

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
  2. Quadratic Formula
  3. Completing the Square

Reminder:

The basic quadratic function equation is:
Y=ax2+bx+cY=ax^2+bx+c

When:
a a   - the coefficient of X2X^2
b b   - the coefficient of XX
cc - the constant term

  • aa must be different from 00
  • bb or cc can be 00
  • a,b,ca,b,c can be negative/positive
  • The quadratic function can also look like this:
    • Y=ax2Y=ax^2
    • Y=ax2+bxY=ax^2+bx
    • Y=ax2+cY=ax^2+c
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Test yourself on solution of quadratic equation!

einstein

What is the value of X in the following equation?

\( X^2+10X+9=0 \)

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How do you solve a quadratic function using trinomial?

Y=ax2+bx+cY=ax^2+bx+c
Find 22 numbers that satisfy the following two conditions:

first numbersecond number=cafirst~number \cdot second~number = c \cdot a
first number+second number=bfirst~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 aca\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 bb.
  3. We will write the pair of numbers that meets the 22 conditions in this form:
    ​​​​​​​0= (x+one solution)(x+second solution)​​​​​​​0=  (x+one~solution)(x+second~solution)
    or with subtraction operations.
  4. 22 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=1a=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:
x29x+14x^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 1414 (and remember the negative numbers as well)
We get:
14,114,1
7,27,2
14,1-14, -1
7,2-7 ,-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 7,2-7 ,-2

We write the factorization:
(x7)(x2)=0(x-7)(x-2)=0
The solutions:
X=7X=7
X=2X=2

How do you solve a quadratic function using the quadratic formula?

Y=ax2+bx+cY=ax^2+bx+c

Meet the quadratic formula:
x=b±b24ac2ax = {-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: 
2x23x+12x^2-3x+1

Let's solve it using the quadratic formula:
First, let's arrange the parameters:
a=2a=2
b=3b=-3
c=1 c=1 

Now, we substitute into the quadratic formula:
For the first time with plus-

(3)+(3)242122=\frac{-(-3)+\sqrt{(-3)^2-4*2*1}}{2*2}=

=3+14=44=1=\frac{3+\sqrt1}4=\frac{4}4=1

The second time with minus:

(3)(3)242122=\frac{-(-3)-\sqrt{(-3)^2-4*2*1}}{2*2}=

=314=24=12=\frac{3-\sqrt1}4=\frac{2}4=\frac{1}2


We got 22 solutions – 
X=12,1X=\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:

a2+2ab+b2=(a+b)2a^2+2ab+b^2=(a+b)^2

a22ab+b2=(ab)2a^2-2ab+b^2=(a-b)^2

Solution method with example:
Here is the function 4x2+8x54x^2+8x-5

  1. Let's look at the quadratic function and focus only on ax2+bxax^2+bx.
    For now, we'll ignore cc. In the example, we'll focus on 4x2+8x4x^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 (ab)2(a-b)^2 or (a+b)2(a+b)^2 as relevant, that will give us what appears in the pair we focused on ax2+bxax^2+bx

In the example-
The appropriate formula for completing the square is:
a2+2ab+b2=(a+b)2a^2+2ab+b^2=(a+b)^2

Let's ask, what should we put as aa and bb to get 4x2+8x4x^2+8x?
The answer is - (2x+2)2(2x+2)^2
Let's expand this expression according to the formula for completing the square and get:
4x2+8x+44x^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 CC from the original function and write it in the equation as well.

    In the example:
    4x2+8x+44x^2+8x+4
    The number 44 was added. To cancel it, we will subtract 44 and not forget the original CC
    4x2+8x45=4x^2+8x-4-5=
  2. We will substitute the pair ax2+bxax^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)245=(2X+2)^2-4-5=
    (2x+2)29(2x+2)^2-9

Now let's set the equation to 0 and solve:
(2x+2)29=0(2x+2)^2-9=0

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

(2x+2)2=32(2x+2)^2=3^2
and also
(2x+2)2=(3)2(2x+2)^2=(-3)^2

Solution 1:
2x+2=32x+2=3
2x=12x=1
x=12x=\frac{1}2

Solution 2: 
2x+2=32x+2=-3
2x=52x=-5
x=212x=-2\frac{1}2

For more information on completing the square, click here!

Do you know what the answer is?
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