Question

If the value of $y=\sqrt{{{x}^{2}}+6x+8}$ then show that one value of $\sqrt{1+iy}+\sqrt{1-iy}=\sqrt{2x+8}$ .

Answer 1

Now we will consider the equation $y=\sqrt{{{x}^{2}}+6x+8}$ . Now we will square the equation and then completing the square on the RHS of the equation we will write the value of x in terms of y. Now we will substitute the obtained value of x in the expression $2x+8$ . Now we will simplify the expression by writing $\sqrt{1+{{y}^{2}}}$ as $\sqrt{1-{{\left( iy \right)}^{2}}}$ and add and subtract the term $iy$ in the expression we will further simplify the expression by using the formula ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$and then take square root to arrive at the required result.

Complete step-by-step answer:
Now consider the given equation $y=\sqrt{{{x}^{2}}+6x+8}$ .
Now squaring both sides we get,
$\begin{aligned} & \Rightarrow {{y}^{2}}={{x}^{2}}+6x+8 \\\ & \Rightarrow {{y}^{2}}={{x}^{2}}+6x+9-1 \\\ \end{aligned}$
Now we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Hence using this we get,
$\Rightarrow {{y}^{2}}={{\left( x+3 \right)}^{2}}-1$
Now rearranging the term we get,
$\Rightarrow {{y}^{2}}+1={{\left( x+3 \right)}^{2}}$
Now again taking square root on both sides we get,
$\Rightarrow \sqrt{{{y}^{2}}+1}=\left( x+3 \right)$
Rearranging the terms in the equation we get,
$\Rightarrow x=\sqrt{1+{{y}^{2}}}-3$
Now let us substitute the value of x in the expression $2x+8$ Hence we get,
$\begin{aligned} & \Rightarrow 2x+8=2\left( \sqrt{1+{{y}^{2}}}-3 \right)+8 \\\ & \Rightarrow 2x+8=2\sqrt{1+{{y}^{2}}}-6+8 \\\ & \Rightarrow 2x+8=2\sqrt{1+{{y}^{2}}}+2 \\\ \end{aligned}$
Now we will rewrite $\sqrt{1+{{y}^{2}}}$ as $\sqrt{1-\left( -{{y}^{2}} \right)}$ and again writing $\left( -{{y}^{2}} \right)={{\left( iy \right)}^{2}}$ as ${{i}^{2}}=-1$ . Hence we get,
$\Rightarrow 2x+8=2\sqrt{1-{{\left( iy \right)}^{2}}}+2$
Now let us add and subtract $iy$ in the expression
$\begin{aligned} & \Rightarrow 2x+8=2\sqrt{1-{{\left( iy \right)}^{2}}}+1+iy+1-iy \\\ & \Rightarrow 2x+8=2\sqrt{1-{{\left( iy \right)}^{2}}}+\left( 1+iy \right)+\left( 1-iy \right) \\\ & \Rightarrow 2x+8=2\sqrt{1-{{\left( iy \right)}^{2}}}+{{\left( \sqrt{1+iy} \right)}^{2}}+{{\left( \sqrt{1-iy} \right)}^{2}} \\\ \end{aligned}$
Now we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ . Hence using this we get,
$\Rightarrow 2x+8={{\left[ \left( \sqrt{1+iy} \right)+\left( \sqrt{1-iy} \right) \right]}^{2}}$
$\Rightarrow \left[ \sqrt{1+iy}+\sqrt{1-iy} \right]=\sqrt{2x+8}$
Hence the given equation is proved.

Note: Now while solving such equation we should try to simplify the equation without square root. Hence taking square roots eliminates the square roots and the equation is easy to solve. Further after simplification we again take the square root and hence arrive at the required equation. Also note that in such problems it helps to solve backwards as well. Hence on the other hand we can start with the equation $\sqrt{1+iy}+\sqrt{1-iy}=\sqrt{2x+8}$ square it and take hints to reach the equation.