Chapter 6.7, Problem 26PPS

To determine

To Solve:

  $\sqrt{x-3}=\sqrt{x+4}-1$

Answer to Problem 26PPS

x = 12

Explanation of Solution

Given Information:

Given equation:

  $\sqrt{x-3}=\sqrt{x+4}-1$

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 $\sqrt{x-3}=\sqrt{x+4}-1$

  $\begin{array}{l}\sqrt{x-3}=\sqrt{x+4}-1\\ \Rightarrow \sqrt{x-3}-\sqrt{x+4}=-\mathrm{1.........................................}[\text{Subtract }\sqrt{x+4}\text{ from each side]}\\ $ \Rightarrow {\left(\sqrt{x-3}-\sqrt{x+4}\right)}^2={(-1)}^2\mathrm{................................}[\text{Squaring each side]}$\Rightarrow x-3+x+4-2\sqrt{x-3}\cdot \sqrt{x+4}=\mathrm{1.................}[Simplify]\\ \Rightarrow 2x+1-2\sqrt{x^2+x-12}=1\\ \Rightarrow -2\sqrt{x^2+x-12}=1-1-2x\mathrm{...........................}[\text{Subtract (1+2x) from each side]}\\ \Rightarrow -2\sqrt{x^2+x-12}=-2x\\ \Rightarrow {\left(-2\sqrt{x^2+x-12}\right)}^2={\left(-2x\right)}^2\mathrm{.......................}[\text{Squaring each side]}\\ \Rightarrow 4(x^2+x-12)=4x^2\\ \Rightarrow 4x^2+4x-48=4x^2\\ \Rightarrow 4x-48=\mathrm{0....................................}[{\text{Subtract 4x}}^{2}from\text{ each side]}\\ \Rightarrow \text{4x=48}\mathrm{............................................}\text{[Add 48 each side]}\\ \Rightarrow \frac{4x}{4}=\frac{48}{4}\mathrm{.................................................}[\text{Divide each side by 4]}\\ \Rightarrow x=12\end{array}$

Check :

  

So, the solution is x = 12