Algebra 2 Common Core

Published by Prentice Hall
ISBN 10: 0133186024
ISBN 13: 978-0-13318-602-4

Chapter 4 - Quadratic Functions and Equations - 4-8 Complex Numbers - Practice and Problem-Solving Exercises - Page 253: 36



Work Step by Step

The given expression, $ x^2+\dfrac{1}{4} ,$ is equivalent to \begin{align*} & x^2-\left(-\dfrac{1}{4}\right) \\\\&= x^2-\left(-1\cdot\dfrac{1}{4}\right) .\end{align*} Since $i^2=-1,$ the expression above is equivalent to \begin{align*} & x^2-\left(i^2\cdot\dfrac{1}{4}\right) \\\\&= x^2-\left(\dfrac{1}{4}i^2\right) .\end{align*} Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of $2$ squares, the expression above is equivalent to \begin{align*} & (x)^2-\left(\dfrac{1}{2}i\right)^2 \\&= \left(x+\dfrac{1}{2}i\right)\left(x-\dfrac{1}{2}i\right) .\end{align*} Hence, the factored form of the given expression is $ \left(x+\dfrac{1}{2}i\right)\left(x-\dfrac{1}{2}i\right) .$ Checking: Multiplying the factors above results to \begin{align*} & \left(x+\dfrac{1}{2}i\right)\left(x-\dfrac{1}{2}i\right) \\\\&= \left(x\right)^2-\left(\dfrac{1}{2}i\right)^2 &\left(\text{use }(a+b)(a-b)=a^2-b^2\right) \\\\&= x^2-\dfrac{1}{4}i^2 \\\\&= x^2-\dfrac{1}{4}(-1) &\left(\text{use }i^2=-1\right) \\\\&= x^2+\dfrac{1}{4} \text{ (same as original expression)} .\end{align*}
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