Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 19: 58


The quadratic formula is $x=\dfrac{-b\pm \sqrt{b^2-4bc}}{2a}$, which can be derived as shown below.

Work Step by Step

The given quadratic equation $ax^2+bx+c=0$ can give quadratic formula as follows: Step 1: Divide throughout by $a$. $\Rightarrow x^2+\dfrac{b}{a}x+\dfrac{c}{a}=0$ Step 2: Add as well as subtract $\left(\dfrac{\text{Coefficient of x}}{2}\right)^2=\left(\dfrac{\frac{b}{a}}{2}\right)^2=\left(\dfrac{b}{2a}\right)^2$. $\Rightarrow x^2+\dfrac{b}{a}x+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2 +\dfrac{c}{a}=0$ Step 3: Multiply and divide the term with $x$ by 2. $\Rightarrow x^2+2\cdot\dfrac{b}{2a}\cdot x+\left(\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2 +\dfrac{c}{a}=0$ Step 4: Compress the perfect square as follows: $\left(x+\dfrac{b}{2a}\right)^2-\left(\dfrac{b}{2a}\right)^2 +\dfrac{c}{a}=0$ Step 5: Now, solve for $x$ as follows: $\Rightarrow \left(x+\dfrac{b}{2a}\right)^2-\dfrac{b^2}{4a^2} +\dfrac{c}{a}=0$ $\Rightarrow\left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2}{4a^2} -\dfrac{c}{a}$ $\Rightarrow\left(x+\dfrac{b}{2a}\right)^2=\dfrac{b^2-4ac}{4a^2}$ $\Rightarrow x+\dfrac{b}{2a}=\pm\dfrac{\sqrt{b^2-4ac}}{2a}$ $\Rightarrow x= -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$ $\Rightarrow x= \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$, which is the required quadratic formula.
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