Question

How do you solve for $x$ in simplest radical form: $2{{\left( x+3 \right)}^{2}}+10=66$?

Answer 1

Hint: We first keep the variable and other constants on one side. We first subtract 10 from both sides. We then divide both sides of the equation by the constant 2. Then we take the square root on both sides of the equation. From that we subtract 3 to the both sides to find the value of $x$ for $2{{\left( x+3 \right)}^{2}}+10=66$.

Complete step by step answer:
We need to find the solution of the given equation $2{{\left( x+3 \right)}^{2}}+10=66$.
We first subtract 10 from both sides of $2{{\left( x+3 \right)}^{2}}+10=66$.
$\begin{align}
  & 2{{\left( x+3 \right)}^{2}}+10-10=66-10 \\
 & \Rightarrow 2{{\left( x+3 \right)}^{2}}=56 \\
\end{align}$
Then we divide both sides of the equation by 10 and get
$\begin{align}
  & \dfrac{2{{\left( x+3 \right)}^{2}}}{2}=\dfrac{56}{2} \\
 & \Rightarrow {{\left( x+3 \right)}^{2}}=28 \\
\end{align}$.
Now we have a quadratic equation ${{\left( x+3 \right)}^{2}}=28$.
We need to find the solution of the given equation ${{\left( x+3 \right)}^{2}}=28$.
We take square root 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.
$\begin{align}
  & \sqrt{{{\left( x+3 \right)}^{2}}}=\sqrt{28} \\
 & \Rightarrow \left( x+3 \right)=\pm 2\sqrt{7} \\
\end{align}$
Now we add $-3$ to the both sides of the equation $\left( x+3 \right)=\pm 2\sqrt{7}$ to get value for variable $x$.
$\begin{align}
  & \left( x+3 \right)-3=-3\pm 2\sqrt{7} \\
 & \Rightarrow x=-3\pm 2\sqrt{7} \\
\end{align}$
The given quadratic equation has two solutions and they are $x=-3\pm 2\sqrt{7}$.

Note:
The highest power of the variable or the degree of a polynomial decides the number of roots or the solution of that polynomial. Quadratic equations have two roots. Cubic polynomials have three. It can be both real and imaginary roots.
We can also apply the quadratic equation formula to solve the equation ${{\left( x+3 \right)}^{2}}=28$.
The simplified form is ${{x}^{2}}+6x-19=0$.
We know for a general equation of quadratic $a{{x}^{2}}+bx+c=0$, the value of the roots of x will be $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
In the given equation we have ${{x}^{2}}+6x-19=0$. The values of a, b, c is $1,6,-19$ respectively.
$x=\dfrac{-6\pm \sqrt{{{6}^{2}}-4\times 1\times \left( -19 \right)}}{2\times 1}=\dfrac{-6\pm \sqrt{112}}{2}=\dfrac{-6\pm 4\sqrt{7}}{2}=-3\pm 2\sqrt{7}$.