Last updated: September 29, 2021

## How to solve quadratics by formula

A quadratic equation is an equation where the highest power in the variable $$x$$(say) is $$2$$ and which can be written in the standard form as

$$ax^{2} +bx+c=0$$, when $$a\neq 0$$

Where $$x$$ represents an unknown variable and $$a$$, $$b$$, $$c$$ represent the constants. Here $$a$$, $$b$$, $$c$$ are the numerical coefficient of the quadratic equation, that is

$$\text{a}$$ represents the numerical coefficient of $$x^{2}$$,

$$\text{b}$$ represents the numerical coefficient of $$x$$,

$$\text{c}$$ represents the constant numerical term.

Here the solution to the quadratic equation is the value of the unknown variable $$x$$ which satisfies the given quadratic equation. The solutions of the quadratic equation are called zero or the root of the equation. For quadratic equation, the equation has two roots or zeros.

## E2.5A: Derive and solve quadratic equation by use of the formula.

Now, we derive the formula which will be used for finding the roots.

Let a quadratic equation be $$ax^{2}+bx+c = 0$$

First, divide both the sides of the above equation by $$a$$.

$$x^{2}+\frac{b}{a}x+\frac{c}{a}=0$$. Now, solve it by completing the square.

$$\left ( x+\frac{b}{2a} \right )^{2}-\left ( \frac{b}{2a} \right )^{2}+\frac{c}{a}=0$$

$$\left ( x+\frac{b}{2a} \right )^{2}-\frac{b^{2}}{4a^{2}}+\frac{c}{a}=0$$

$$\left ( x+\frac{b}{2a} \right )^{2}-\left ( \frac{b^{2}-4ac}{4a^{2}} \right )=0$$

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

Now taking a square root to solve it further.

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

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

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

Therefore, $$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$$.

Therefore, the root of the quadratic equations are $$\frac{-b+ \sqrt{b^{2}-4ac}}{2a}$$ or $$\frac{-b - \sqrt{b^{2}-4ac}}{2a}$$.

If $$b^{2}-4ac=0$$. Then the equation $$ax^{2}+bx+c=0$$ will have no real root.

So, by using the formula $$x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}$$, we can find the roots of the quadratic equation. Here the values of $$x$$ represents the roots of the equation.

### Worked examples of solving quadratics by formula

Example 1: Find the root of the quadratic equation $$x^{2}+2x+1=0$$.

Step 1: Compare the above equation with the equation $$ax^{2}+bx+c=0$$ to get the value of $$a$$, $$b$$ and $$c$$.

From the given equation, $$a=1$$, $$b=2$$ and $$c=1$$.

Step 2: Recall the formula of roots of the quadratic equation.

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

Step 3: Substitute the values of $$a$$, $$b$$ and $$c$$ to the above formula.

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

$$\frac{-2\pm \sqrt{2^{2}-4\times 1\times 1}}{2\times 1}$$

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

Step 4: Answer in the preferred notation.

Therefore, $$x=-1$$, $$x=-1$$ are the roots of the quadratic equation.

Example 2: Find the root of the quadratic equation $$x^{2}+6x+5=0$$.

Step 1: Compare the above equation with the equation $$ax^{2}+bx+c=0$$ to get the value of $$a$$, $$b$$ and $$c$$.

From the given equation, $$a=1$$, $$b=6$$ and $$c=5$$.

Step 2: Recall the formula of the roots of the quadratic equation.

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

Step 3: Substitute the values of $$a$$, $$b$$ and $$c$$ in the above formula.

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

$$\frac{-6\pm \sqrt{6^{2}-4\times 1\times 5}}{2\times 1}$$

$$\frac{-6\pm \sqrt{36-20}}{2}\\=\frac{-6\pm \sqrt{16}}{2}$$

$$\frac{-6\pm 4}{2}$$

Therefore, $$x=\frac{-6+4}{2}=\frac{-2}{2}=-1$$ and $$x=\frac{-6-4}{2}=\frac{-10}{2}=-5$$

Step 4: Answer in the preferred notation.

Therefore $$x=-1$$ and $$x=-5$$ are the roots of the given quadratic equation.

Example 3: Solve the equation $$\left ( 4x-3 \right )^{2}-2\left ( x+3 \right )=0$$ and find the roots of the equation.

Step 1: Simplify the above equation.

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

$$16x^{2}-24x+9-2x-6=0$$

$$16x^{2}-26x+3=0$$

Step 2: Compare the above equation with the equation $$ax^{2}+bx+c=0$$ to get the value of $$a$$, $$b$$ and $$c$$.

From the given equation, $$a=16$$, $$b=-26$$ and $$c=3$$.

Step 3: Recall the formula of the roots of the quadratic equation.

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

Step 4: Substitute the values of $$a$$, $$b$$ and $$c$$ in the above equation.

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

$$\frac{26\pm \sqrt{26^{2}-4\times 16\times 3}}{2\times 16}$$

$$\frac{26\pm \sqrt{676-192}}{32}$$

\frac{26\pm \sqrt{484}}{32}=\frac{26\pm 22}{32}\end{align*}.

Therefore, $$x=\frac{26+22}{32}=\frac{48}{32}=\frac{3}{2}$$ and $$x=\frac{26-22}{32}=\frac{4}{32}=\frac{1}{8}$$.

Step 5: Answer in the preferred notation.

Therefore $$x=\frac{3}{2}$$ and $$x=\frac{1}{8}$$ are the roots of the given quadratic equation.