Question

In the following exercises, solve.

$$\begin{gathered} \left(a\right)\sqrt{u}+1=\sqrt{u+4} \\ \left(b\right)\sqrt{n-5}+4=\sqrt{3n+7} \end{gathered}$$

Short answer

Part (a) : The solutions for $\sqrt{u}+1=\sqrt{u+4}$is $\frac{9}{4}$

Part (b) : The solution for $\sqrt{n-5}+4=\sqrt{3n+7}$are 14 and 6

Step-by-step solution

Step 1. Given information 

The given expressions are

$$\begin{gathered} \left(a\right)\sqrt{u}+1=\sqrt{u+4} \\ \left(b\right)\sqrt{n-5}+4=\sqrt{3n+7} \end{gathered}$$

Step 2. Squaring on both sides 

$$\begin{gathered} \left(a\right)\sqrt{u}+1=\sqrt{u+4} \\ {\left(\sqrt{u}+1\right)}^2={\left(\sqrt{u+4}\right)}^2 \\ u+2\sqrt{u}+1=u+4 \\ 2\sqrt{u}=3 \\ squaringonbothsides \\ 4u=9 \\ u=\frac{9}{4} \end{gathered}$$

$$\begin{gathered} \left(b\right)\sqrt{n-5}+4=\sqrt{3n+7} \\ squaringonbothsides \\ {\left(\sqrt{n-5}+4\right)}^2={\left(\sqrt{3n+7}\right)}^2 \\ {\left(\sqrt{n-5}\right)}^2+4^2+8\sqrt{n-5}=3n+7 \\ n-5+16+8\sqrt{n-5}=3n+7 \\ 8\sqrt{n-5}=2n-4 \\ Againsquaringonbothsides \\ {\left(8\sqrt{n-5}\right)}^2={\left(2n-4\right)}^2 \\ 64\left(n-5\right)=4n^2-16n+16 \\ 64n-320=4n^2-16n+16 \\ 4n^2-80n+336=0 \\ n=14n=6 \end{gathered}$$

The solutions are 14 and 6

Step 3. The solution 

The solution for $\sqrt{u}+1=\sqrt{u+4}$is $\frac{9}{4}$

The solution for$\sqrt{n-5}+4=\sqrt{3n+7}$are 14 and 6