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Question: How do you solve \({{\left( x+2 \right)}^{2}}=-7\) ?...

How do you solve (x+2)2=−7{{\left( x+2 \right)}^{2}}=-7 ?

Explanation

Solution

We have to find the solution of (x+2)2=−7{{\left( x+2 \right)}^{2}}=-7. Then we take the square root on both sides of the equation. The right-hand side being negative, we get imaginary value as the square root value. From that we subtract 2 to the both sides to find the value of xx for (x+2)2=−7{{\left( x+2 \right)}^{2}}=-7. Then we put the solution value in the equation to verify the result.

Complete step-by-step solution:
We need to find the solution of the given equation (x+2)2=−7{{\left( x+2 \right)}^{2}}=-7.
We take square roots on both sides of the equation. As the equation is a quadratic one, the number of roots will be 2 and they are equal in value but opposite in sign.
(x+2)2=−7=±i7 ⇒(x+2)=±i7 \begin{aligned} & \sqrt{{{\left( x+2 \right)}^{2}}}=\sqrt{-7}=\pm i\sqrt{7} \\\ & \Rightarrow \left( x+2 \right)=\pm i\sqrt{7} \\\ \end{aligned}
Here ii is the complex value.
Now we subtract 2 to the both sides of the equation (x+2)=±i7\left( x+2 \right)=\pm i\sqrt{7} to get value for variable xx.
(x+2)−2=±i7−2 ⇒x=−2±i7 \begin{aligned} & \left( x+2 \right)-2=\pm i\sqrt{7}-2 \\\ & \Rightarrow x=-2\pm i\sqrt{7} \\\ \end{aligned}
The given quadratic equation has two solutions and they are x=−2±i7x=-2\pm i\sqrt{7}.

Note: We try to verify the value of the root of x=−2±i7x=-2\pm i\sqrt{7} for the equation (x+2)2=−7{{\left( x+2 \right)}^{2}}=-7.
Putting the value x=−2+i7x=-2+i\sqrt{7} in the left side of the equation we get

& {{\left( x+2 \right)}^{2}} \\\ & ={{\left( -2+i\sqrt{7}+2 \right)}^{2}} \\\ & ={{\left( i\sqrt{7} \right)}^{2}} \\\ & =7\times {{i}^{2}} \\\ & =7\times \left( -1 \right) \\\ & =-7 \\\ \end{aligned}$$ Therefore, the value $x=-2+i\sqrt{7}$ satisfies the equation ${{\left( x+2 \right)}^{2}}=-7$ as $f\left( -2+i\sqrt{7} \right)=0$. Same thing can be said for $x=-2-i\sqrt{7}$.