Question

How do you solve the equation ${{x}^{2}}-2x-10=0$ by completing the square?

Answer 1

Hint: To solve the equation ${{x}^{2}}-2x-10=0$ by completing the square. First, we have to change the equation so that the constant term in a quadratic equation $a{{x}^{2}}+bx+c=0$ to the right side. That is c is moved to the right side. Now, we have to check whether the coefficient of ${{x}^{2}}$, that is, a is equal to 1 or not. If a is not equal to 1, we have to divide both the sides by a so that we get the coefficient of ${{x}^{2}}$ as 1. Next, we have to add both sides by the square of half of the coefficient of x, that is ${{\left( \dfrac{b}{2a} \right)}^{2}}$ . Now, let us consider the LHS and make it as a square of binomial. Then, we have to take the square root of both the sides and solve for x.

Complete step by step solution:
We need to solve the equation ${{x}^{2}}-2x-10=0$ by completing the square. First, we have to change the equation so that the constant term in a quadratic equation $a{{x}^{2}}+bx+c=0$ to the right side. That is c is moved to the right side. We can do the same for ${{x}^{2}}-2x-10=0$ .
$\Rightarrow {{x}^{2}}-2x=10$
Now, we have to check whether the coefficient of ${{x}^{2}}$, that is, a is equal to 1 or not. If a is not equal to 1, we have to divide both the sides by a so that we get the coefficient of ${{x}^{2}}$ as 1. In this problem, we can skip this step as we have a=1.
Next, we have to add both sides by the square of half of the coefficient of x, that is ${{\left( \dfrac{b}{2a} \right)}^{2}}$ . This means, we have to add ${{\left( \dfrac{-2}{-2} \right)}^{2}}={{1}^{2}}=1$ on both sides. We will get
$\begin{align}
  & {{x}^{2}}-2x+1=10+1 \\
 & \Rightarrow {{x}^{2}}-2x+1=11 \\
\end{align}$
Now, let us consider the LHS and make it as a square of binomial. We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ .
$\begin{align}
  & \Rightarrow {{x}^{2}}-2x+1=11 \\
 & \Rightarrow {{\left( x-1 \right)}^{2}}=11 \\
\end{align}$
Now, we have to take the square root of both the sides.
$\begin{align}
  & \Rightarrow x-1=\pm \sqrt{11} \\
 & \Rightarrow x=1\pm \sqrt{11} \\
\end{align}$
Hence, the answer is $x=1\pm \sqrt{11}$ .

Note: Students must be thorough with all the steps. For equations with a not equal to 1, we must divide both the sides by a so that we get the coefficient of ${{x}^{2}}$ as 1. Students must be careful when taking the square root of both the sides. They must never miss the $\pm $ sign.