Question

Solve the following equation by completing the square method, \[{{m}^{2}}-3m-1=0\] .

Answer 1

Hint: We have the equation \[{{m}^{2}}-3m-1=0\] which is quadratic in m. Now, transform this quadratic equation as \[\left( {{m}^{2}}-2.m.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}-1=0 \right)\] and then use the formula \[{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] to simplify it. Now, apply square root and then solve it further.

Complete step-by-step answer:
According to the question, it is given that our equation is \[{{m}^{2}}-3m-1=0\] and we have to solve this equation by using the perfect square method.
\[{{m}^{2}}-3m-1=0\] ……………………….(1)
The given equation is quadratic in m.
We have to make the quadratic equation a perfect square of some linear equation in m.
Transforming equation (1), we get
\[{{m}^{2}}-3m-1=0\]
\[\begin{align}
  & \Rightarrow {{(m)}^{2}}-2.m.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}-1=0 \\
 & \Rightarrow {{(m)}^{2}}-2.m.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}={{\left( \dfrac{3}{2} \right)}^{2}}+1 \\
 & \Rightarrow {{(m)}^{2}}-2.m.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}=\dfrac{9}{4}+1 \\
\end{align}\]
\[\Rightarrow {{(m)}^{2}}-2.m.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}=\dfrac{13}{4}\] ………………………(2)
We know the formula,
\[{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] ………………………..(3)
Replacing a by m and b by \[\dfrac{3}{2}\] in equation (3), we get
\[{{\left( m-\dfrac{3}{2} \right)}^{2}}={{m}^{2}}-2.m.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}\] ………………..(4)
We can see that (2) can be further simplified. Since both equation (2) and equation (4) has \[{{m}^{2}}-2.m.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}\] , so in equation (2) we can replace \[{{m}^{2}}-2.m.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}\] by \[{{\left( m-\dfrac{3}{2} \right)}^{2}}\] .
Now, transforming equation (2) and replacing \[{{m}^{2}}-2.m.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}\] by \[{{\left( m-\dfrac{3}{2} \right)}^{2}}\] , we get
\[\begin{align}
  & \Rightarrow {{(m)}^{2}}-2.m.\dfrac{3}{2}+{{\left( \dfrac{3}{2} \right)}^{2}}=\dfrac{13}{4} \\
 & \Rightarrow {{\left( m-\dfrac{3}{2} \right)}^{2}}=\dfrac{13}{4} \\
\end{align}\]
Applying Square root on both LHS and RHS of the above equation, we get
\[\Rightarrow {{\left\{ {{\left( m-\dfrac{3}{2} \right)}^{2}} \right\}}^{\dfrac{1}{2}}}={{\left( \dfrac{13}{4} \right)}^{\dfrac{1}{2}}}\] …………………..(5)
We know the formula, \[{{({{x}^{m}})}^{n}}={{x}^{mn}}\] .
Replace x by \[\left( m-\dfrac{3}{2} \right)\] , m by 2, and n by \[\dfrac{1}{2}\] in the above formula, we get
\[\Rightarrow {{\left\{ {{\left( m-\dfrac{3}{2} \right)}^{2}} \right\}}^{\dfrac{1}{2}}}=\left( m-\dfrac{3}{2} \right)\] ………………..(6)
Using equation (6) for simplifying equation (5), we get
\[\Rightarrow {{\left\{ {{\left( m-\dfrac{3}{2} \right)}^{2}} \right\}}^{\dfrac{1}{2}}}={{\left( \dfrac{13}{4} \right)}^{\dfrac{1}{2}}}\]
\[\Rightarrow \left( m-\dfrac{3}{2} \right)=\pm \dfrac{\sqrt{13}}{2}\]
The above equation is linear in m and it can be solved easily. Taking the constant term \[\dfrac{3}{2}\] to the RHS, we get
\[\Rightarrow m=\dfrac{3}{2}\pm \dfrac{\sqrt{13}}{2}\] .
Hence, the value of m is \[\dfrac{3\pm \sqrt{13}}{2}\] .

Note: In this question, one may write the quadratic equation \[{{m}^{2}}-3m-1=0\] as \[{{m}^{2}}+3m-1=0\] and then think to convert the equation \[{{m}^{2}}+3m-1=0\] into the form of \[{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\] . After converting and solving it we will get \[x=\dfrac{-3\pm \sqrt{13}}{2}\] . This is completely wrong. We can see that our answer is completely different. So, we have to avoid this type of silly mistakes.