Algebra 2 Common Core

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

Chapter 4 - Quadratic Functions and Equations - 4-7 The Quadratic Formula - Practice and Problem-Solving Exercises - Page 245: 36

Answer

$\begin{align*} \text{Discriminant: }& 0 \\\text{Number of Real Solutions: }& 1 \end{align*}$

Work Step by Step

Using the properties of equality, the given equation, $ 12x(x+1)=-3 ,$ is equivalent to \begin{align*} 12x(x)+12x(1)&=-3 &\text{ (use Distributive Property)} \\ 12x^2+12x&=-3 \\ 12x^2+12x+3=0 \end{align*} In the equation above $a= 12 ,$ $b= 12 ,$ and $c= 3 .$ Using the Discriminant Formula which is given by $b^2-4ac,$ the value of the discriminant is \begin{array}{l}\require{cancel} & 12^2-4(12)(3) \\&= 144-144 \\&= 0 \end{array} Since the discriminant is equal to $0,$ then there is $1$ real solution. Hence, \begin{align*} \text{Discriminant: }& 0 \\\text{Number of Real Solutions: }& 1 \end{align*}
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