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Chapter 1: Problem 116

In solving \(\sqrt{3 x+4}-\sqrt{2 x+4}=2,\) why is it a good idea to isolate a radical term? What if we don't do this and simply square each side? Describe what happens.

Short Answer

Expert verified
The solution to the given equation is \(x = 8 \pm 4\sqrt{3}\). Isolating a radical term before squaring can simplify the process of solution and mitigate the risk of introducing errors or extraneous solutions. If we don't isolate the radical term, we'll introduce additional terms involving a square root, making the problem more complex to resolve. Hence it's advisable to isolate the radical term before squaring both sides in a square root equation.

Step by step solution

01

Isolate a radical term

Firstly, isolate \(\sqrt{3x+4}\) on one side of the equation by adding \(\sqrt{2x+4}\) to both sides: \(\sqrt{3x+4}=2+\sqrt{2x+4}\).
02

Square both sides

Next, square both sides in order to eliminate the square root on the left side. \((\sqrt{3x+4})^2 = (2 + \sqrt{2x+4})^2\) which simplifies to \(3x+4 = 4 + 4\sqrt{2x+4}+ 2x+4\).
03

Simplify the equation

Simplify the equation by grouping the like terms and move the radical term to one side: \(x = 4\sqrt{2x+4}\). Squaring both sides again will give, \(x^2 =16x+64\). After rearranging the expression, we get \(x^2 -16x - 64 = 0\).
04

Solve the quadratic equation

Solve for \(x\) by using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), which will give \(x = 8 \pm 4\sqrt{3}\).
05

Check for Extraneous solutions

Substitute the solution in the original equation and make sure they're valid. In this case, both solutions are valid.

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