Question

How do you complete the square to solve \[{{x}^{2}}-2x-3=0\]?

Answer 1

Hint: In this problem, we have to solve the given quadratic equation using the complete square method. We have to make the LHS side as a perfect square. So first we will take the constant term to the RHS side.
Then we have to make the LHS side as a perfect square by doing required operations. Then we will get a perfect square to solve for x.

Complete step-by-step answer:
So the given equation is
\[{{x}^{2}}-2x-3=0\]
Now we can add 3 on both sides in the above equation, we get
\[\begin{align}
  & \Rightarrow {{x}^{2}}-2x-3+3=0+3 \\
 & \Rightarrow {{x}^{2}}-2x=3 \\
\end{align}\]
Now we have to make it a perfect square on the LHS side.
we already know that
\[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
So to convert the equation into this form we have to add 1 on both sides of the equation.
we get
\[\Rightarrow {{x}^{2}}-2x+1=3+1\]
We will get
\[\Rightarrow {{x}^{2}}-2x+1=4\]
We have already discussed the formula of perfect square.
Now we can apply this formula in the left-hand side of the above equation, we get
\[\Rightarrow {{\left( x-1 \right)}^{2}}=4\]
We can take square on both sides, we get
\[\Rightarrow \sqrt{{{\left( x-1 \right)}^{2}}}=\pm \sqrt{4}\]
We can cancel the square root on the left-hand side,
\[\Rightarrow x-1=\pm \sqrt{4}\]
Now we know that the square root of \[4\] is \[2\]. We will get
\[\Rightarrow x-1=\pm 2\]
Now we can add the number 1 on both sides, we get
\[\Rightarrow x=1\pm 2\]
We will get
\[\Rightarrow x=3,x=-1\]
Therefore, the value of \[x=3,-1\] .

Note: We can also do this by different methods but here it is given that we have to do it using complete square method otherwise we can use quadratic formula. While doing this method so also students can use direct formula to add \[{{\left( \frac{b}{2} \right)}^{2}}\]and then we can solve.