How to Solve Quadratic Equations

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Using the Quadratic Formula to solve quadratic equations

To use the quadratic formula, we must change the quadratic equation we are solving into "standard form". If we don't do this, none of the following steps will work. The goal is to change the quadratic equation so that the quadratic expression is on one side of the equation and only the number zero, $0$ , is on the other. Look at the equation below.

$ax^2 \ + \ bx \ + \ c \ = \ 0$

This format makes it easy to find out what $a, \ b,$ and $c$ 's numerical values are! Once we know these values, we can plug them into the quadratic formula and figure out the values of $x$ .

The formula for quadratic equations

$ax^2 \ + \ bx \ + \ c \ = \ 0 \ ⇒ \ x \ = \ \frac{-b \ ± \ \sqrt{b^2 \ - \ 4ac}}{2a}$

Where $a, \ b,$ and $c$ are the coefficients of any quadratic equation in the standard form, $ax^2 \ + \ bx \ + \ c \ = \ 0$ .

Simplify Be careful at each step to make the expressions simpler. Students often make mistakes because they tend to "chill out," which can lead to errors that could have been avoided, like when adding, subtracting, multiplying, or dividing real numbers.

Example:

Solve $2x^2 \ + \ 3x \ - \ 2 \ = \ 0$ .

Solution:

Given,

$2x^2 \ + \ 3x \ - \ 2 \ = \ 0 \ ⇒ \ a \ = \ 2, \ b \ = \ 3, \ c \ = \ -2$

We know that $b^2 \ - \ 4ac \ = \ 3^2 \ - \ 4(2)(-2) \ = \ 9 \ - \ (-16) \ = \ 25 \ > \ 0$ .

So, the roots exist. So, $x \ = \ \frac{-b \ ± \ \sqrt{b^2 \ - \ 4ac}}{2a} \ = \ \frac{-3 \ ± \ \sqrt{25}}{4}$

i.e., $x_1 \ = \ \frac{-3 \ - \ 5}{4} \ = \ -\frac{8}{4} \ = \ -2$ and $x_2 \ = \ \frac{-3 \ + \ 5}{4} \ = \ \frac{2}{4} \ = \ \frac{1}{2}$

So, the answer to $2x^2 \ + \ 3x \ - \ 2 \ = \ 0$ is either $-2$ or $\frac{1}{2}$ .

Solving Quadratic Equations By Factorisation

The quadratic equation can be written as the product of factors whose degree is less than or equal to two. It's called "factoring the quadratic equation," a way to solve quadratic equations.

Let's see an example to learn.

Example:

Use the factoring method to solve $2x^2 \ + \ 6x \ + \ 4 \ = 0$ .

Solution:

$2x^2 \ + \ 6x \ + \ 4 \ = 0 \ ⇒ \ 2x^2 \ + \ 2x \ + \ 4x \ + \ 4 \ = \ 0 \ ⇒ \ 2x(x \ + \ 1) \ + \ 4(x \ + \ 1) \ = 0$ $⇒ \ (x \ + \ 1)(2x \ + \ 4) \ = 0$ $⇒ \ x \ + \ 1 \ = 0$ or $2x \ + \ 4 \ = 0$ :

$x \ + \ 1 \ = 0 \ ⇒ \ x \ = \ -1$
$2x \ + \ 4 \ = 0 \ ⇒ \ 2x \ = \ -4 \ ⇒ \ x \ = \ -2$

So, the answer to the given equation is either $x \ = \ -1$ or $x \ = \ -2$ .

Free printable Worksheets

Exercises for Solving Quadratic Equations

1) Solve: $x^2 \ + \ 9x \ + \ 20 \ = \ 0$

2) Solve: $x^2 \ + \ 2x \ - \ 3 \ = \ 0$

3) Solve: $2x^2 \ + \ 7x \ - \ 4 \ = \ 0$

4) Solve: $2x^2 \ + \ 4x \ + \ 2 \ = \ 0$

5) Solve: $18x^2 \ - \ 13x \ + \ 2 \ = \ 0$

6) Solve: $5x^2 \ + \ 27x \ - \ 18 \ = \ 0$

7) Solve: $2x^2 \ + \ 7x \ + \ 3 \ = \ 0$

8) Solve: $-6x^2 \ + \ 29x \ - \ 35 \ = \ 0$

9) Solve: $4x^2 \ + \ 8x \ + \ 3 \ = \ 0$

10) Solve: $x^2 \ + \ 3x \ - \ 4 \ = \ 0$

Answers

1) Solve: $x^2 \ + \ 9x \ + \ 20 \ = \ 0$

$\color{red}{x^2 \ + \ 2x \ - \ 3 \ = \ 0 \ ⇒ \ (x \ + \ 4)(x \ + \ 5) \ = \ 0}$

$\color{red}{(x \ + \ 4) \ = \ 0 \ ⇒ \ x \ = \ -4}$
$\color{red}{(x \ + \ 5) \ = \ 0 \ ⇒ \ x \ = \ -5}$

$\color{red}{x_1 \ = \ -4 \ , \ x_2 \ = \ -5}$

2) Solve: $x^2 \ + \ 2x \ - \ 3 \ = \ 0$

$\color{red}{x^2 \ + \ 9x \ + \ 20 \ = \ 0 \ ⇒ \ a \ = \ 1, \ b \ = \ 2, \ c \ = \ -3}$
$\color{red}{Δ \ = \ b^2 \ - \ 4ac \ = \ 2^2 \ - \ 4(1)(-3) \ = \ 4 \ - \ (-12) \ = \ 16}$
$\color{red}{x \ = \ \frac{-b \ ± \ \sqrt{Δ}}{2a} \ = \ \frac{-2 \ ± \ \sqrt{16}}{2} \ ⇒}$ $\color{red}{x_1 \ = \ \frac{-2 \ - \ 4}{2} \ = \ -\frac{6}{2} \ = \ -3 \ , \ }$ $\color{red}{x_2 \ = \ \frac{-2 \ + \ 4}{2} \ = \ \frac{2}{2} \ = \ 1}$

3) Solve: $2x^2 \ + \ 7x \ - \ 4 \ = \ 0$

$\color{red}{x^2 \ + \ 9x \ + \ 20 \ = \ 0 \ ⇒ \ a \ = \ 2, \ b \ = \ 7, \ c \ = \ -4}$
$\color{red}{Δ \ = \ b^2 \ - \ 4ac \ = \ 7^2 \ - \ 4(2)(-4) \ = \ 49 \ - \ (-32) \ = \ 81}$
$\color{red}{x \ = \ \frac{-b \ ± \ \sqrt{Δ}}{2a} \ = \ \frac{-7 \ ± \ \sqrt{81}}{4} \ ⇒}$ $\color{red}{x_1 \ = \ \frac{-7 \ - \ 9}{4} \ = \ -\frac{16}{4} \ = \ -4 \ , \ }$ $\color{red}{x_2 \ = \ \frac{-7 \ + \ 9}{4} \ = \ \frac{2}{4} \ = \ \frac{1}{2}}$

4) Solve: $2x^2 \ + \ 4x \ + \ 2 \ = \ 0$

$\color{red}{2x^2 \ + \ 4x \ + \ 2 \ = \ 0 \ ⇒ \ 2x^2 \ + \ 2x \ + \ 2x \ + \ 2 \ = \ 0 \ ⇒ \ 2x(x \ + \ 1) \ + \ 2(x \ + \ 1) \ = \ 0}$ $\color{red}{⇒ \ (x \ + \ 1)(2x \ + \ 2) \ = \ 0}$

$\color{red}{(x \ + \ 1) \ = \ 0 \ ⇒ \ x \ = \ -1}$
$\color{red}{(2x \ + \ 2) \ = \ 0 \ ⇒ \ x \ = \ -1}$

$\color{red}{x_1 \ = \ x_2 \ = \ -1}$

5) Solve: $18x^2 \ - \ 13x \ + \ 2 \ = \ 0$

$\color{red}{18x^2 \ - \ 13x \ + \ 2 \ = \ 0 \ ⇒ \ a \ = \ 18, \ b \ = \ -13, \ c \ = \ 2}$
$\color{red}{Δ \ = \ b^2 \ - \ 4ac \ = \ (-13)^2 \ - \ 4(18)(2) \ = \ 169 \ - \ 144 \ = \ 25}$
$\color{red}{x \ = \ \frac{-b \ ± \ \sqrt{Δ}}{2a} \ = \ \frac{13 \ ± \ \sqrt{25}}{36} \ ⇒}$ $\color{red}{x_1 \ = \ \frac{13 \ - \ 5}{36} \ = \ \frac{8}{36} \ = \ \frac{2}{9} \ , \ }$ $\color{red}{x_2 \ = \ \frac{13 \ + \ 5}{36} \ = \ \frac{18}{36} \ = \ \frac{1}{2}}$

6) Solve: $5x^2 \ + \ 27x \ - \ 18 \ = \ 0$

$\color{red}{5x^2 \ + \ 27x \ - \ 18 \ = \ 0 \ ⇒ \ a \ = \ 5, \ b \ = \ 27, \ c \ = \ -18}$
$\color{red}{Δ \ = \ b^2 \ - \ 4ac \ = \ (27)^2 \ - \ 4(5)(-18) \ = \ 729 \ - \ (-360) \ = \ 1089}$
$\color{red}{x \ = \ \frac{-b \ ± \ \sqrt{Δ}}{2a} \ = \ \frac{-27 \ ± \ \sqrt{1089}}{10} \ ⇒}$ $\color{red}{x_1 \ = \ \frac{-27 \ - \ 33}{10} \ = \ \frac{60}{10} \ = \ 6 \ , \ }$ $\color{red}{x_2 \ = \ \frac{-27 \ + \ 33}{10} \ = \ \frac{6}{10} \ = \ 0.6}$

7) Solve: $2x^2 \ + \ 7x \ + \ 3 \ = \ 0$

$\color{red}{2x^2 \ + \ 7x \ + \ 3 \ = \ 0 \ ⇒ \ 2x^2 \ + \ 6x \ + \ x \ + \ 3 \ = \ 0 \ ⇒ \ 2x(x \ + \ 3) \ + \ 1(x \ + \ 3) \ = \ 0}$ $\color{red}{⇒ \ (2x \ + \ 1)(x \ + \ 3) \ = \ 0}$

$\color{red}{(2x \ + \ 1) \ = \ 0 \ ⇒ \ x \ = \ -\frac{1}{2}}$
$\color{red}{(x \ + \ 3) \ = \ 0 \ ⇒ \ x \ = \ -3}$

$\color{red}{x_1 \ = \ -\frac{1}{2} \ , \ x_2 \ = \ -3}$

8) Solve: $-6x^2 \ + \ 29x \ - \ 35 \ = \ 0$

$\color{red}{-6x^2 \ + \ 29x \ - \ 35 \ = \ 0 \ ⇒ \ -6x^2 \ + \ 14x \ + \ 15x \ - \ 35 \ = \ 0}$ $\color{red}{ \ ⇒ \ -2x(3x \ - \ 7) \ + \ 5(3x \ - \ 7) \ = \ 0⇒ \ (-2x \ + \ 5)(3x \ - \ 7) \ = \ 0}$

$\color{red}{(-2x \ + \ 5) \ = \ 0 \ ⇒ \ x \ = \ \frac{5}{2} \ = \ 2.5}$
$\color{red}{(3x \ - \ 7) \ = \ 0 \ ⇒ \ x \ = \ \frac{7}{3}}$

$\color{red}{x_1 \ = \ 2.5 \ , \ x_2 \ = \ \frac{7}{3}}$

9) Solve: $4x^2 \ + \ 8x \ + \ 3 \ = \ 0$

$\color{red}{4x^2 \ + \ 8x \ + \ 3 \ = \ 0 \ ⇒ \ a \ = \ 4, \ b \ = \ 8, \ c \ = \ 3}$
$\color{red}{Δ \ = \ b^2 \ - \ 4ac \ = \ (8)^2 \ - \ 4(4)(3) \ = \ 64 \ - \ 48 \ = \ 16}$
$\color{red}{x \ = \ \frac{-b \ ± \ \sqrt{Δ}}{2a} \ = \ \frac{-8 \ ± \ \sqrt{16}}{8} \ ⇒}$ $\color{red}{x_1 \ = \ \frac{-8 \ - \ 4}{8} \ = \ -\frac{12}{8} \ = \ -\frac{3}{2} \ , \ }$ $\color{red}{x_2 \ = \ \frac{-8 \ + \ 4}{8} \ = \ -\frac{4}{8} \ = \ -\frac{1}{2}}$

10) Solve: $x^2 \ + \ 3x \ - \ 4 \ = \ 0$

$\color{red}{x^2 \ + \ 3x \ - \ 4 \ = \ 0 \ ⇒ \ a \ = \ 1, \ b \ = \ 3, \ c \ = \ -4}$
$\color{red}{Δ \ = \ b^2 \ - \ 4ac \ = \ (3)^2 \ - \ 4(1)(-4) \ = \ 9 \ + \ 16 \ = \ 25}$
$\color{red}{x \ = \ \frac{-b \ ± \ \sqrt{Δ}}{2a} \ = \ \frac{-3 \ ± \ \sqrt{25}}{2} \ ⇒}$ $\color{red}{x_1 \ = \ \frac{-3 \ - \ 5}{2} \ = \ -\frac{9}{2} \ = \ -4.5 \ , \ }$ $\color{red}{x_2 \ = \ \frac{-3 \ + \ 5}{2} \ = \ \frac{2}{2} \ = \ 1}$

How to Solve Quadratic Equations