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 - Page 502: 46


The solution is $12$.

Work Step by Step

Subtract $6$ and add $\sqrt{3x}$ to both sides of the equation to obtain: $$x-\sqrt{3x}-6+\sqrt{3x}=6-6+\sqrt{3x} \\x-6=\sqrt{3x}$$ Square both sides. Note that $(a-b)^2=a^2-2ab+b^2$. $$(x-6)^2=(\sqrt{3x})^2 \\x^2-2(x)(6)+6^2=3x \\x^2-12x+36=3x \\x^2-12x+36-3x=0 \\x^2-15x+36=0$$ Factor the trinomial: $$(x-12)(x-3)=0$$ Use the Zero Factor Property by equating each factor to zero, then solve each equation to obtain: \begin{array}{ccc} \\&x-12=0 &\text{or} &x-3=0 \\&x=12 &\text{or} &x=3 \end{array} Note that $3$ is does not satisfy the original equation so it is not a solution. Therefore, the solution is $12$.
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