Question

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Solve: $\sqrt{n-4}+5=\sqrt{3n+3}$

Short answer

Ans: $$\begin{gathered} \sqrt{n-4}+5=\sqrt{3n+3} \\ \Rightarrow n=\frac{15\sqrt{5}+43}{2},\frac{15\sqrt{5}-43}{2} \end{gathered}$$

Step-by-step solution

Step 1. Given information.

given expression $\sqrt{n-4}+5=\sqrt{3n+3}$

Step 2. Square both sides of the equation and solve it.   

$$\begin{gathered} {(\sqrt{n-4}+5)}^2={\left(\sqrt{3n+3}\right)}^2 \\ \Rightarrow n-4+25+10\sqrt{n-4}=3n+3 \\ \Rightarrow 10\sqrt{n-4}=2n-18 \\ \Rightarrow 5\sqrt{n-4}=n-9 \end{gathered}$$

Step 3. 

$$\begin{gathered} {(5\sqrt{n-4})}^2={(n-9)}^2 \\ \Rightarrow 25(n-4)=n^2+81-18n \\ \Rightarrow n^2-43n+181=0 \end{gathered}$$

Step 4. Find the Discriminant of the equation n2−43n+181=0 .

Discriminant $D=b^2-4ac$

here, $a=1,b=-43,c=181$

$D=-{43}^2-4\times 1\times 181=1125$

Since, Discriminant is positive then roots are real and unequal.

Step 5. Then the value of n.

$$\begin{gathered} n=\frac{-b+\sqrt{D}}{2a},\frac{-b-\sqrt{D}}{2a} \\ \Rightarrow n=\frac{-(-43)+\sqrt{1125}}{2\times 1},\frac{-(-43)-\sqrt{1125}}{2\times 1} \\ \Rightarrow n=\frac{15\sqrt{5}+43}{2},\frac{15\sqrt{5}-43}{2} \end{gathered}$$