Question

Solve the following quadratic equation:
${{m}^{2}}=2+2m$

Answer 1

Hint: The equation given in the above question is a quadratic equation in m. We can solve the given equation ${{m}^{2}}-2m-2=0$ by the Sridharacharya formula/Quadratic formula of finding roots of a quadratic equation $a{{m}^{2}}+bm+c=0$ which is $m=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step-by-step solution -
The equation given in the question which we have to solve is:
$\begin{align}
  & {{m}^{2}}=2+2m \\
 & \Rightarrow {{m}^{2}}-2m-2=0 \\
\end{align}$
The above equation is a quadratic equation in m.
There are various ways to solve the quadratic equation. Here, we are solving through the Sridharacharya formula of finding the roots of a quadratic equation $a{{m}^{2}}+bm+c=0$ which is shown below.
$m=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Comparing the quadratic equation ${{m}^{2}}-2m-2=0$ with $a{{m}^{2}}+bm+c=0$ for finding the roots of the given equation using Sridharacharya rule we get,
After comparing the given quadratic equation ${{m}^{2}}-2m-2=0$ with $a{{m}^{2}}+bm+c=0$, the values of a, b and c are as follows:
$a=1,b=-2,c=-2$
Substituting these values of a, b and c in the Sridharacharya formula of finding the roots of a quadratic equation we get,
$m=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
$\begin{align}
  & \Rightarrow m=\dfrac{2\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( 1 \right)\left( -2 \right)}}{2} \\
 & \Rightarrow m=\dfrac{2\pm \sqrt{4+8}}{2} \\
 & \Rightarrow m=\dfrac{2\pm \sqrt{12}}{2} \\
 & \Rightarrow m=\dfrac{2\pm 2\sqrt{3}}{2}=1\pm \sqrt{3} \\
\end{align}$
From the above calculations, the values of m that we have got are:
$\begin{align}
  & m=1+\sqrt{3}; \\
 & m=1-\sqrt{3} \\
\end{align}$
Hence, the values of m after solving the given equation are $m=1+\sqrt{3},m=1-\sqrt{3}$.

Note: You can solve this quadratic equation in the following way:
 ${{m}^{2}}=2+2m$
Rearranging the above quadratic equation we get,
${{m}^{2}}-2m-2=0$
We can make the part of the above quadratic equation as a perfect square by adding and subtracting 1 on the left hand side of the above equation.
${{m}^{2}}-2m+1-1-2=0$
We know that ${{\left( m-1 \right)}^{2}}={{m}^{2}}-2m+1$ so using this relation in the above equation we get,
${{\left( m-1 \right)}^{2}}-3=0$
Rewriting the above equation in the form of ${{a}^{2}}-{{b}^{2}}$ we get,
${{\left( m-1 \right)}^{2}}-{{\left( \sqrt{3} \right)}^{2}}=0$
We know the algebraic identity that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. As you can see the above relation is in the form of ${{a}^{2}}-{{b}^{2}}$ so we can use this identity.
$\left( m-1+\sqrt{3} \right)\left( m-1-\sqrt{3} \right)=0$
Equating each bracket in the above equation equal to 0 we get,
$\begin{align}
  & m-1+\sqrt{3}=0 \\
 & \Rightarrow m=1-\sqrt{3} \\
 & m-1-\sqrt{3}=0 \\
 & \Rightarrow m=1+\sqrt{3} \\
\end{align}$
Hence, we have got the value of $m=1\pm \sqrt{3}$ which is the same as we have got above.