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# Solving Quadratic Equations – Formulas, Tricks & Examples

November 3, 2022

This post is also available in: हिन्दी (Hindi)

Quadratic equations are the polynomial equations of degree $2$ in one variable of the form $p\left(x \right) = ax^{2} + bx + c = 0$ where $a$, $b$, $c$ are real numbers and $a \ne 0$. The quadratic equations have two zeroes or roots which can be real numbers, imaginary numbers, or both.

There are various methods of solving quadratic equations. The most popular method is solving quadratic equations by factoring. Let’s understand these different methods of solving quadratic equations with examples.

## Methods of Solving Quadratic Equations

Solving an equation is a process of finding the zeroes(or roots) of an equation. The zeroes (or roots) are also known as solutions of the equation. These are the values of a variable that satisfies the equation.

Since a quadratic equation is $2$-degree equation, it can have a maximum of $2$ real roots, i.e., a quadratic equation can have any one of the following

• zero real root
• one real root (or two equal real roots)
• two real roots (or two unequal real roots)

Note: The terms ‘no real root’ or ‘one real root’ do not mean that a quadratic equation can have $0$ or $1$ roots. A quadratic equation always has $2$ roots, where

• both are real
• one is real and one is imaginary
• both are imaginary

There are different ways of solving quadratic equations. The most common methods are:

• Solving quadratic equations by graphing
• Solving quadratic equations by factoring
• Solving quadratic equations by completing the square

Apart from these methods, there are some other methods that are used only in specific cases (when the quadratic equation has missing terms) which are also discussed in the later sections.

Maths can be really interesting for kids

### Solving Quadratic Equations by Graphing

$x$-intercept(s) of an equation are the points, where its curve crosses the $x$-axis. The $x$-coordinate of these points are the values of $x$ for which $y$ is $0$. In other words, we can say that the $x$-coordinate of $x$-intercepts are the zeroes or the roots of an equation.

Thus to find the roots or zeroes of any equation, we can graph that equation and note down its $x$-intercepts.

For solving the quadratic equations by graphing, we first have to graph the quadratic expression (when the equation is in the standard form) either manually or by using a graphing calculator. Then the $x$-intercept(s) of the graph is noted.

These are the steps to solve quadratic equations by a graphing method.

Step 1: Express the equation in the form $y = ax^{2} + bx + c$

Step 2: Graph the quadratic expression

Step 3: Identify the $x$-intercepts

Step 4: The roots of the quadratic equation are the $x$-intercepts

### Examples

Ex 1: Solve the quadratic equation $x^{2} + 10 = 7x$ by graphing.

Writing the equation in standard form

$x^{2} + 10 = 7x => x^{2} – 7x + 10 = 0$

Now, expressing it in the form of a function $f\left(x \right) = x^{2} – 7x + 10$ and plotting its graph

From the above image we that the $x$-intercepts of the quadratic equation $x^{2} + 10 = 7x$ are $\left( 2, 0 \right)$ and $\left(5, 0 \right)$, therefore, the roots of the quadratic equation $x^{2} + 10 = 7x$ are $2$ and $5$.

Note: The quadratic equation $x^{2} + 10 = 7x$ has two real unequal roots.

Ex 2: Solve the quadratic equation $x^{2} – 8x + 16$ by graphing.

Expressing the given equation in the form of a function $f\left(x \right) = x^{2} – 8x + 16$ and plotting its graph.

From the above image we that the $x$-intercept of the quadratic equation $x^{2} – 8x + 16$ is $\left( 4, 0 \right)$, therefore, the root of the quadratic equation $x^{2} + 10 = 7x$ is $4$.

Note: The quadratic equation $x^{2} + 10 = 7x$ has two real equal roots.

Ex 3: Solve the quadratic equation $2x^{2} – 3x + 3$ by graphing.

Expressing the given equation in the form of a function $f\left(x \right) = 2x^{2} – 3x + 3$ and plotting its graph.

From the above image we that there are no $x$-intercepts of the quadratic equation $2x^{2} – 3x + 3$, therefore, the quadratic equation $x^{2} + 10 = 7x$ is $4$ has no real roots.

### Solving Quadratic Equations by Factoring

Solving quadratic equations by factoring is one of the popular methods used to solve quadratic equations. These are the steps to solve quadratic equations by the factoring method.

Step 1: Get the equation into standard form. i.e., get all the terms to one side (usually to the left side) of the equation such that the other side is $0$

Step 2: Factor the quadratic expression

Step 3: Set each of the factors to zero

Step 4: Solve each of the equations obtained

Examples

Ex 1: Solve the quadratic equation $x^{2} – 2x – 24$ by factoring method.

$x^{2} – 2x – 24 = 0$

Factoring the LHS of the equation, we get

$x^{2} + 4x – 6x – 24 = 0$

$=>x\left(x + 4 \right) – 6\left(x + 4 \right) = 0$

$=>\left(x + 4 \right) \left(x – 6 \right) = 0$

$=>x + 4 = 0 \text{ or } x – 6 = 0$

$=>x = -4 \text{ or } x = 6$

Therefore, the roots of the quadratic equation $x^{2} – 2x – 24$ are $-4$ and $6$.

Note: The quadratic equation $x^{2} – 2x – 24$ has two real equal roots.

Ex 2: Solve the quadratic equation $x^{2} – 10x + 25$ by factoring method.

$x^{2} – 10x + 25$

Factoring the LHS of the equation, we get

$x^{2} – 5x – 5x + 25 = 0$

$=>x\left(x – 5 \right) -5\left(x – 5\right) = 0$

$=>\left(x – 5\right) \left(x – 5 \right) = 0$

$=>\left(x – 5\right)^{2} = 0$

$=>x – 5 = 0 => x = 5$

Therefore, the root of the quadratic equation $x^{2} – 10x + 25$ is $5$.

Note: The quadratic equation $x^{2} – 10x + 25$ has two real equal roots.

### Solving Quadratic Equations by Completing the Square

Completing the square means writing the quadratic expression $ax^{2} + bx + c$ into the form $a \left(x – h \right)^{2} + k$ (which is also known as vertex form), where $h = -\frac{b}{2a}$ and $k$ can be obtained by substituting $x = h$ in $ax^{2} + bx + c$.  These are the steps to solve quadratic equations by the factoring method.

Step 1: Get the equation into standard form.

Step 2: Complete the square on the left side.

Step 3: Solve it for $x$ by taking the square root on both sides.

### Examples

Ex 1: Solve the quadratic equation $x^{2} + 5x – 7$ by completing the square method.

$x^{2} + 5x – 7 = 0$

$=>x^{2} + 2 \times x \times \frac{5}{2} – 7 = 0$

$=>x^{2} + 2 \times x \times \frac{5}{2} + \left( \frac{5}{2}\right)^{2} – \left( \frac{5}{2}\right)^{2} – 7 = 0$

(Add and subtracting $+ \left( \frac{5}{2}\right)^{2}$)

$=>\left(x^{2} + 2 \times x \times \frac{5}{2} + \left( \frac{5}{2}\right)^{2} \right) – \left( \frac{5}{2}\right)^{2} – 7 = 0$

$=>\left(x + \frac{5}{2} \right)^{2} – \frac{53}{4} = 0$

$=>\left(x + \frac{5}{2} \right)^{2} = \frac{53}{4}$

Taking the square root of both sides

$=>x + \frac{5}{2} = \pm \sqrt{\frac{53}{4}}$

$=>x + \frac{5}{2} = \pm \frac{\sqrt{53}}{2}$

$=>x = – \frac{5}{2} \pm \frac{\sqrt{53}}{2}$

$=> x = \frac{5 \pm \sqrt{53}}{2}$

Therefore, $x = \frac{5 + \sqrt{53}}{2}$, $x = \frac{5 – \sqrt{53}}{2}$

Ex 2: Solve the quadratic equation $x^{2} – 2x – 24$ by completing the square method.

$x^{2} – 2x – 24 = 0$

$=>x^{2} – 2 \times x \times 1 – 24 = 0$

$=>x^{2} – 2 \times x \times 1 + 1^{2} – 1^{2} – 24 = 0$

$=>\left(x^{2} – 2 \times x \times 1 + 1^{2} \right) – 1 – 24 = 0$

$=>\left(x – 1\right)^{2} – 25 = 0$

$=>\left(x – 1\right)^{2} = 25$

Taking the square root of both sides

$x – 1 = \pm \sqrt{25}$

$=> x – 1 = \pm 5$

$=> x = 1 \pm 5$

$=> x = 1 + 5$ or $=> x = 1 – 5$

$=> x = 6$ or $=> x = -4$

Ex 3: Solve the quadratic equation $2x^{2} + 8x + 3 = 0$ using completing the square method.

$2x^{2} + 8x + 3 = 0$

$=>2\left(x^{2} + 4x \right) + 3 = 0$

$=>2\left(x^{2} + 2 \times x \times 2\right) + 3 = 0$

$=>2\left(x^{2} + 2 \times x \times 2 + 2^{2} – 2^{2}\right) + 3 = 0$

$=>2\left(x^{2} + 2 \times x \times 2 + 2^{2} \right) – 2 \times 2^{2} + 3 = 0$

$=>2\left(x + 2\right)^{2} – 8 + 3 = 0$

$=>2\left(x + 2\right)^{2} – 5 = 0$

$=>2\left(x + 2\right)^{2} = 5$

$=>\left(x + 2\right)^{2} = \frac {5}{2}$

Taking the square root of both sides

$=>x + 2 = \pm \sqrt{\frac {5}{2}}$

$=>x + 2 = \pm \frac{\sqrt{5}}{\sqrt{2}}$

$=>x + 2 = \pm \frac{\sqrt{10}}{2}$

$=>x = -2 \pm \frac{\sqrt{10}}{2}$

$=>x = -2 + \frac{\sqrt{10}}{2}$, $=>x = -2 – \frac{\sqrt{10}}{2}$

$=> x = \frac{-4 + \sqrt{10}}{2}$, $=> x = \frac{-4 – \sqrt{10}}{2}$

As seen above the other methods for solving the quadratic equations have some limitations such as the factoring method is useful only when the quadratic expression is factorable, the graphing method being useful only when the quadratic equation has real roots, etc.

But solving quadratic equations by quadratic formula overcomes all these limitations and is useful to solve any type of quadratic equation.  These are the steps to solve quadratic equations by the quadratic formula.

Step 1: Get the given equation to standard form.

Step 2: Compare the equation with $ax^{2} + bx + c = 0$ and find the values of $a$, $b$, and $c$.

Step 3: Substitute the values into the quadratic formula which says $x = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}$.

Step 4: Simplify and get the values of $x$

The standard form of a quadratic equation is $ax^{2} + bx + c = 0$.

Dividing both sides by $a$, we get $x^{2} + \frac {b}{a} x + \frac{c}{a} = 0$

$=>x^{2} + 2 \times x \times \frac {b}{2a} x + \frac{c}{a} = 0$

$=>x^{2} + 2 \times x \times \frac {b}{2a} = – \frac{c}{a}$

Adding and subtracting by $\left(\frac {b}{2a} \right)^{2}$

$=>x^{2} + 2 \times x \times \frac {b}{2a} + \left(\frac {b}{2a} \right)^{2} = – \frac{c}{a} + \left(\frac {b}{2a} \right)^{2}$

$=> \left(x + \frac{b}{2a} \right)^{2} = \frac{b^{2} – 4ac}{4a^{2}}$

Taking the square root of both sides

$=> x + \frac{b}{2a} = \pm \frac{\sqrt{b^{2} – 4ac}}{2a}$

$=> x = – \frac{b}{2a} \pm \frac{\sqrt{b^{2} – 4ac}}{2a}$

$=> x = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}$, and hence the quadratic formula.

Note: $b^{2} – 4ac$ is called the discriminant of the quadratic equation and is denoted by $\text{D}$ or $\triangle$.

### Examples

Ex 1: Solve the quadratic equation $2x^{2} + 9x + 7$ by using the quadratic formula.

Comparing $2x^{2} + 9x + 7$ with the standard form of the quadratic equation $ax^{2} + bx + c$, we get $a = 2$, $b = 9$, and $c = 7$

The quadratic formula is $x = \frac{-b \pm \sqrt{b^{2} – 4ac}}{2a}$

Substituting the values of $a$, $b$, and $c$ in the formula we get

$x = \frac{-9 \pm \sqrt{9^{2} – 4 \times 2 \times 7}}{2 \times 2}$

$=>x = \frac{-9 \pm \sqrt{81 – 56}}{4}$

$=>x = \frac{-9 \pm \sqrt{25}}{4}$

$=>x = \frac{-9 \pm 5}{4}$

$=>x = \frac{-9 + 5}{4}$, $=>x = \frac{-9 – 5}{4}$

$=>x = \frac{-4}{4}$, $=>x = \frac{-14}{4}$

$=>x = -1$, $=>x = -\frac{7}{2}$

## How to Solve a Bi Quadratic Equation?

A bi quadratic equation also known as a quartic equation is an equation of the form $ax^{4} + bx^{3} + cx^{2} + dx + e = 0$. The degree of such equations is $4$.

One of the ways of solving a bi quadratic equation is to reduce it to a quadratic equation and then use any of the above-mentioned methods to solve the equation.

Let’s consider an example of solving a bi quadratic equation.

### Examples

Ex 1: Solve $x^{4} – 2x^{2} – 24$

$x^{4} – 2x^{2} – 24$ is a bi quadratic equation as the degree of the equation is $4$

The given equation can be written as $\left(x^{2} \right)^{2} – 2x^{2} – 24$

Replacing $x^{2}$ by $u$, we get $u^{2} – 2u – 24$

$u^{2} – 2u – 24$ is a quadratic equation.

Let’s solve this equation by the factoring method.

$u^{2} – 2u – 24 = 0$

$=>u^{2} – 6u + 4u – 24 = 0$

$=>u \left(u – 6 \right) + 4 \left(u – 6 \right) = 0$

$=>\left(u – 6 \right) \left(u + 4 \right) = 0$

$=> u – 6 = 0$ or $u + 4 = 0$

$=> u = 6$ or $u = -4$

When $u = 6$

$x^{2} = 6 => x = \pm \sqrt{6} => x = \sqrt{6}$, $x = -\sqrt{6}$

When $u = -4$

$x^{2} = -4 => x = \pm \sqrt{-4}$ which are imaginary values

Therefore, $x^{4} – 2x^{2} – 24$ has two real roots: $x = \sqrt{6}$, $x = -\sqrt{6}$

## Practice Problems

1. Solve the following quadratic equations using the graphing method
• $x^{2} – 2x + 35$
• $x^{2} – 18x + 81$
• $2x^{2} + x + 5$
2. Solve the following quadratic equations using the factoring method
• $2x^{2} – x – 6$
• $3x^{2} – 7x – 6$
3. Solve the following quadratic equations using the completing the square method
• $3x^{2} + 5x – 1$
• $x^{2} – 4x + 3$
• $x^{2} – 7x + 9$
• $3x^{2} + 12x + 7$

## FAQs

### What is the meaning of solving quadratic equations?

Solving quadratic equations means finding their solutions or roots. i.e., it is the process of finding the values of the variable that satisfy the equation.

### What are the four different ways of solving quadratic equations?

The four ways of solving the quadratic equations are
a) Solving quadratic equations by graphing
b) Solving quadratic equations by factoring
c) Solving quadratic equations by completing the square