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Chapter 1: Problem 25

Solve each radical equation in Check all proposed solutions. $$ \sqrt{2 x+3}+\sqrt{x-2}=2 $$

Short Answer

Expert verified
The solution to the radical equation is \(x = 3\). The proposed solution \(x = 11\) is extraneous and is not a solution of the original equation.

Step by step solution

01

Isolate the first square root term

Start by isolating \( \sqrt{2 x+3} \) on one side of the equation by subtracting \( \sqrt{x-2} \) from both sides of the equation. This gives the equation: \[ \sqrt{2 x+3} = 2 - \sqrt{x-2} \]
02

Square both sides to eliminate the first square root term

If both sides of the above equation are squared, the square root on the left-hand side is eliminated: \[(\sqrt{2 x+3})^2 = (2 - \sqrt{x-2})^2 \]This simplifies to:\[2x + 3 = 4 - 4 \sqrt{x-2} + x - 2\]Then, collect like terms and rearrange to get a single square root term on one side of the equation:\[4 \sqrt{x-2} = x - 1\]
03

Square both sides again to eliminate the second square root term

If both sides of the above equation are squared, the remaining square root is eliminated: \[(4 \sqrt{x-2})^2 = (x - 1)^2\]This simplifies to:\[16(x - 2) = x^2 - 2x + 1\]Rearrange the equation and simplify terms to get a quadratic equation:\[x^2 - 18x + 33 = 0\]
04

Solve the quadratic equation

Finally, the quadratic equation can be solved using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Using this formula, the solutions for x can be found to be \(x = 3\) and \(x = 11\).
05

Check the proposed solutions

Plug the solutions \(x = 3\) and \(x = 11\) back into the original radical equation to check for validity.For \(x = 3\),\[ \sqrt{2(3)+3}+\sqrt{(3)-2} = 2\]This simplifies to \(\sqrt{9} + \sqrt{1} = 2\) which is true, hence solution \(x = 3\) is valid.For \(x = 11\),\[ \sqrt{2(11)+3}+\sqrt{(11)-2} = 2\]This simplifies to \(\sqrt{25} + \sqrt{9} = 2\), which is false, hence solution \(x = 11\) is not valid.

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