Solving Quadratic Equations Using the abc Formula

Aan Triono included in category Grade 9

2020-08-24 2026-08-08 215 words One minute

Contents

Introduction

To find the roots of a quadratic equation, we can use factoring, completing the square, or the quadratic formula often called the abc formula:

x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

This article focuses on solving quadratic equations using that formula.

Quadratic formula or abc formula

Find the roots of:

2x^2+x-6=0

From the equation, we get:

Use the abc formula:

\begin{aligned} x_{1,2} &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{-1 \pm \sqrt{1^2 - 4(2)(-6)}}{2(2)} \\ &= \frac{-1 \pm \sqrt{1+48}}{4} \\ &= \frac{-1 \pm \sqrt{49}}{4} \\ &= \frac{-1 \pm 7}{4} \end{aligned}

So:

\begin{aligned} x_1 &= \frac{-1+7}{4} = \frac{6}{4} = \frac{3}{2} \\ x_2 &= \frac{-1-7}{4} = \frac{-8}{4} = -2 \end{aligned}

Therefore, the roots are:

x=\frac{3}{2} \quad \text{and} \quad x=-2

Find the roots of:

2x^2+13x-7=0

From the equation, we get:

Use the abc formula:

\begin{aligned} x_{1,2} &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{-13 \pm \sqrt{13^2 - 4(2)(-7)}}{2(2)} \\ &= \frac{-13 \pm \sqrt{169+56}}{4} \\ &= \frac{-13 \pm \sqrt{225}}{4} \\ &= \frac{-13 \pm 15}{4} \end{aligned}

So:

\begin{aligned} x_1 &= \frac{-13+15}{4} = \frac{2}{4} = \frac{1}{2} \\ x_2 &= \frac{-13-15}{4} = \frac{-28}{4} = -7 \end{aligned}

Therefore, the roots are:

x=\frac{1}{2} \quad \text{and} \quad x=-7

That is how quadratic equations can be solved using the quadratic formula or abc formula.