Chapter 7.7, Problem 1BCYP

To determine

To Solve:

  $\sqrt{x+15}=5+\sqrt{x}$

Answer to Problem 1BCYP

No solution

Explanation of Solution

Given Information:

Given equation:

  $\sqrt{x+15}=5+\sqrt{x}$

Concept Used :

Solve Radical equations:

• Isolate the radical on one side of the equation.
• Raise each side of the equation to a power equal to the index of the radical to eliminate the radical .
• Solve the resulting polynomial equation.Check your results.
• When solving radical equations, the result may be a number that does not satisfy the original equation . Such a number is called an extraneous solution.

Calculation:

• Given $\sqrt{x+15}=5+\sqrt{x}$

  $\begin{array}{l}\sqrt{x+15}=5+\sqrt{x}\\ \Rightarrow {\left(\sqrt{x+15}\right)}^2={\left(5+\sqrt{x}\right)}^2\mathrm{......................}[\text{Squaring each side]}\\ \Rightarrow x+15=25+x+10\sqrt{x}\mathrm{.............................}[{\left(a+b\right)}^2=a^2+b^2+2ab]\\ \Rightarrow x+15-x-25=25+x+10\sqrt{x}-x-\mathrm{25..................}[subtract\text{ (x+25) from each side]}\\ \Rightarrow -10=10\sqrt{x}\\ \Rightarrow \sqrt{x}=-\mathrm{1........................................}[\text{divide each side by 10]}\\ \text{Since the index is 2, which is even ; so the radical will be non- negative}\text{.}\\ \text{But here, it is -1 ; hence, there is no solution}\text{.}\end{array}$