Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 6 - Algebra: Equations and Inequalities - 6.5 Quadratic Equations - Exercise Set 6.5 - Page 400: 46

Answer

the solution set is\[\left\{ 7 \right\}\].

Work Step by Step

Step 1:Shift all nonzero terms to left side and obtain zero on the other side. For this, add 49 both sides as follows: \[{{x}^{2}}-14x+49=-49+49\] This implies that, \[{{x}^{2}}-14x+49=0\] So, after shifting all nonzero terms to the left side, the above equation becomes: \[{{x}^{2}}-14x+49=0\] Step 2:Find the factor of the above equation: Consider the equation\[{{x}^{2}}-14x+49=0\]. Factorize it as follows: \[\begin{align} & {{x}^{2}}-14x+49=0 \\ & {{x}^{2}}-7x-7x+49=0 \\ & x\left( x-7 \right)-7\left( x-7 \right)=0 \\ & \left( x-7 \right)\left( x-7 \right)=0 \end{align}\] Steps 3 and 4: Set each factor equal to zero and solve the resulting equation: From step 2, \[\left( x-7 \right)\left( x-7 \right)=0\]. By the zero product principal, \[\left( x-7 \right)=0\]. That is,\[x=7\]repeated twice. Step 5: Check the solution in the original equation. Check for\[x=7\]. So consider, \[\begin{align} & {{x}^{2}}-14x+49=0 \\ & {{\left( 7 \right)}^{2}}-14\left( 7 \right)+49=0 \\ & 49-98+49=0 \\ & 0=0 \end{align}\] Hence, the solution set is\[\left\{ 7 \right\}\].
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