Question

In the following exercises, solve:

(a) $\sqrt{a}+2=\sqrt{a+4}$

(b)$\sqrt{b-2}+1=\sqrt{3b+2}$.

Short answer

The solution is,

Part (a) : $a=0$

Part (b) : no possible value

Step-by-step solution

Part (a), Step 1. Given Information.

The expression is,

$\sqrt{a}+2=\sqrt{a+4}$

Part (a). Step 2. Finding the solution.

$\sqrt{a}+2=\sqrt{a+4}$

Squaring both sides,

role="math" localid="1647933714265" $$\begin{gathered} \Rightarrow {(\sqrt{a}+2)}^2={\left(\sqrt{a+4}\right)}^2 \\ \Rightarrow a+4+4\sqrt{a}=a+4 \end{gathered}$$

Subtracting $4$from both sides,

$$\begin{gathered} \Rightarrow a+4\sqrt{a}=a+4-4 \\ \Rightarrow a+4\sqrt{a}=a \end{gathered}$$

Subtracting $a$from both sides,

$$\begin{gathered} \Rightarrow 4\sqrt{a}=a-a \\ \Rightarrow 4\sqrt{a}=0 \end{gathered}$$

Dividing both sides by $4$,

$$\begin{gathered} \Rightarrow \frac{4\sqrt{a}}{4}=\frac{0}{4} \\ \Rightarrow \sqrt{a}=0 \end{gathered}$$

Squaring both sides,

$$\begin{gathered} \Rightarrow {\left(\sqrt{a}\right)}^2=0^2 \\ \Rightarrow a=0 \end{gathered}$$

Part (a). Step 3. Checking the solution.

$\sqrt{a}+2=\sqrt{0}+2$

$$\begin{gathered} =0+2 \\ =2 \end{gathered}$$

$\sqrt{a+4}=\sqrt{0+4}$

$$\begin{gathered} =\sqrt{4} \\ =2 \end{gathered}$$

The solution is true.

Part (b). Step 1. Given Information.

The expression is,

$\sqrt{b-2}+1=\sqrt{3b+2}$.

Part (b). Step 2. Finding the solution.

$\sqrt{b-2}+1=\sqrt{3b+2}$

Squaring on both sides,

localid="1647934260062" $$\begin{gathered} \Rightarrow {(\sqrt{b-2}+1)}^2={\left(\sqrt{3b+2}\right)}^2 \\ \Rightarrow b-2+1+2\sqrt{b-2}=3b+2 \\ \Rightarrow b-1+2\sqrt{b-2}=3b+2 \end{gathered}$$

Adding localid="1647934268440" $1$on both sides,

$$\begin{gathered} \Rightarrow b+2\sqrt{b-2}=3b+2+1 \\ \Rightarrow b+2\sqrt{b-2}=3b+3 \end{gathered}$$

Subtracting $b$from both sides,

$$\begin{gathered} \Rightarrow 2\sqrt{b-2}=3b-b+3 \\ \Rightarrow 2\sqrt{b-2}=2b+3 \end{gathered}$$

Dividing both sides by $2$,

$$\begin{gathered} \Rightarrow \frac{2\sqrt{b-2}}{2}=\frac{2b+3}{2} \\ \Rightarrow \sqrt{b-2}=b+\frac{3}{2} \end{gathered}$$

Squaring on both sides,

localid="1647934500602" $$\begin{gathered} \Rightarrow {\left(\sqrt{b-2}\right)}^2={(b+\frac{3}{2})}^2 \\ \Rightarrow b-2=b^2+\frac{9}{4}+3b \end{gathered}$$

Adding $2$on both sides,

$$\begin{gathered} \Rightarrow b=b^2+\frac{9}{4}+2+3b \\ \Rightarrow b=b^2+\frac{17}{4}+3b \end{gathered}$$

Subtracting $b$from both sides,

$$\begin{gathered} b^2+\frac{17}{4}+3b-b=0 \\ {b}^2+2b+\frac{17}{4}=0 \\ 4b^2+8b+17=0 \end{gathered}$$

Splitting the middle term such that the sum of the terms is $8$and the product of the terms is $68$. This is not possible. Hence no solution.