Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 8 - Section 8.3 - Solving Equations by Using Quadratic Methods - Exercise Set: 45

Answer

The solution is $4$.

Work Step by Step

Subtract $2$ and add $\sqrt{x}$ to both sides of the equation to obtain: $$x-\sqrt{x}-2+\sqrt{x}=2-2+\sqrt{x} \\x-2=\sqrt{x}$$ Square both sides. Note that $(a-b)^2=a^2-2ab+b^2$. $$(x-2)^2=(\sqrt{x})^2 \\x^2-2(x)(2)+2^2=x \\x^2-4x+4=x \\x^2-4x+4-x=0 \\x^2-5x+4=0$$ Factor the trinomial: $$(x-4)(x-1)=0$$ Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} \\&x-4=0 &\text{or} &x-1=0 \\&x=4 &\text{or} &x=1 \end{array} Note that $1$ is does not satisfy the original equation so it is not a solution. Therefore, the solution is $4$.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.