Question

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

Answer 1

Hint: In this problem, we have been given a quadratic equation and we are asked to solve the given quadratic equation by completing the square. Under this method, find the coefficient of x, find its half and then square it. Add the resultant term to both the sides of the equation. After this, form the square and you will have your answer.

Complete step by step answer:
Given quadratic equation is ${x^2} + 10x - 2 = 0$
Now add $2$ to both sides of the given quadratic equation. We get,
${x^2} + 10x - 2 + 2 = 0 + 2$ , here $ + 2$ and $ - 2$ get cancelled by each other,
$ \Rightarrow {x^2} + 10x = 2$ …. (1)
Divide the coefficient of the $x$ term by $2$ and square the result.
${\left( {\dfrac{{10}}{2}} \right)^2} = {5^2} = 25$
Add it to both sides of the equation (1), we get,
$ \Rightarrow {x^2} + 10x + 25 = 2 + 25$ , adding the numbers in the right-hand side, we get
$ \Rightarrow {x^2} + 10x + 25 = 27$
Rewriting the above equation as –
${x^2} + (2 \times 5)x + 25 = 27$ …. (2)
Now, we have a perfect square trinomial on the left-hand side of the above equation with the form ${a^2} + {b^2} + 2ab = {\left( {a + b} \right)^2}$ , now compare this equation with the equation (2), we get
$a = x,b = 5$ and ${\left( {x + 5} \right)^2} = 27$
Take the square root on both sides, we get,
$\left( {x + 5} \right) = \pm \sqrt {27} $
$ \Rightarrow \left( {x + 5} \right) = \pm \sqrt {3 \times 9} $ , square root of $9$ is $3$
$ \Rightarrow \left( {x + 5} \right) = \pm 3\sqrt 3 $
Now subtract $5$ from both sides, we get
$x + 5 - 5 = \pm 3\sqrt 3 - 5$ , here in the left-hand side both $ - 5$ and $ + 5$ get cancelled by each other, we get,
$x = \pm 3\sqrt 3 - 5$

There are two values of $x$ they are $x = + 3\sqrt 3 - 5$ and $x = - 3\sqrt 3 - 5$.

Note: Completing the square is the process of adding a term to convert the given quadratic expression into something that factors as the square of a binomial. That is we have completed the expression to create a perfect square binomial. In this problem, we have added ${5^2}$ to equation (1) and then we converted equation (2) to the square of the binomial and then we found the values of $x$ .