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Chapter 1: Problem 19

Solve each radical equation in Check all proposed solutions. $$ \sqrt{3 x}+10=x+4 $$

Short Answer

Expert verified
The solution to the equation \( \sqrt{3x} + 10 = x + 4\) is \( x = 12 \)

Step by step solution

01

Isolate the square root

Isolate the square root term on one side of the equation. This can be done by subtracting 4 from both sides of the equation, getting: \[ \sqrt{3x} = x - 6 \]
02

Square both sides

Square both sides of the equation to eliminate the square root on the left side. This gives: \[ 3x = (x - 6)^2 = x^2 - 12x + 36 \]
03

Rearrange the equation

Rearrange the resulting quadratic equation to the form \(ax^2+bx+c=0\). To do this, subtract \(3x\) from both sides, getting: \[x^2 - 15x + 36 = 0\]
04

Solve the quadratic equation

Solve the quadratic equation for \(x\) by using the quadratic formula \(x = {-b \pm \sqrt{b^2-4ac}} / {2a}\). So: \[x = \frac{15 \pm \sqrt{(-15)^2 - 4(1)(36)}}{2} = \frac{15 \pm \sqrt{225 - 144}}{2} = \frac{15 \pm \sqrt{81}}{2}\]Therefore the roots are \(x = \frac{15 \pm 9}{2}\), which simplifies to \(x = 3\) or \(x = 12\)
05

Check the solutions

Check the solutions by substituting them back into the original equation to make sure they satisfy it: For \(x = 3\),\[ \sqrt{3(3)}+10 = 3+4; 7 \neq 7\]Hence, \(x=3\) is not a solution.For \(x = 12\), \[ \sqrt{3(12)}+10 = 12+4; 16 = 16\]So, \(x=12\) is a valid solution.

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Most popular questions from this chapter

Solve each rational inequality in Exercises \(29-48,\) and graph the solution set on a real number line. Express each solution set in interval notation. $$ \frac{x+4}{x}>0 $$

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