Question

How do you solve using the completing the square method ${x^2} - 5x = 9$?

Answer 1

Hint: In this question, we need to solve the given equation using completing the square method. Here we are given a quadratic equation. We need to find the value of the variable x. Firstly, we add ${\left( {\dfrac{5}{2}} \right)^2}$ to both the sides of the quadratic equation and then make rearrangement to make it as a complete square using the mathematical operations. Then we try to solve it to obtain the required solution.

Complete step by step answer:
Given the equation ${x^2} - 5x = 9$
We have to solve the given quadratic equation by completing the square method.
Let us consider the quadratic equation ${x^2} - 5x = 9$ …… (1)
First of all we need to make the coefficient of ${x^2} = 1$. Already the coefficient of ${x^2}$ is 1, so no need to make any changes in it.
So we proceed further to get the solution.
Now we add the square of half the coefficient of x term on both sides of the above equation to complete it as square and we solve for the variable x.
In this case we add ${\left( {\dfrac{{ - 5}}{2}} \right)^2}$which is same as ${\left( {\dfrac{5}{2}} \right)^2}$.
So adding ${\left( {\dfrac{{ - 5}}{2}} \right)^2}$ to both sides of the equation (1), we get,
$ \Rightarrow {x^2} - 5x + {\left( {\dfrac{5}{2}} \right)^2} = 9 + {\left( {\dfrac{5}{2}} \right)^2}$
$ \Rightarrow {x^2} - 5x + {\left( {\dfrac{5}{2}} \right)^2} = 9 + \dfrac{{25}}{4}$
Taking LCM on the R.H.S. we get,
$ \Rightarrow {x^2} - 5x + {\left( {\dfrac{5}{2}} \right)^2} = \dfrac{{36 + 25}}{4}$
$ \Rightarrow {x^2} - 5x + {\left( {\dfrac{5}{2}} \right)^2} = \dfrac{{61}}{4}$
We simplify the middle term in the L.H.S. and rewrite the equation as,
$ \Rightarrow {x^2} - 2x\left( {\dfrac{5}{2}} \right) + {\left( {\dfrac{5}{2}} \right)^2} = \dfrac{{61}}{4}$ …… (2)
Note that the L.H.S. of the above equation is of the form ${a^2} - 2ab + {b^2}$.
We have the algebraic formula given by ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.
Here $a = x,$ $b = \dfrac{5}{2}$.
Therefore ${x^2} - 2x\left( {\dfrac{5}{2}} \right) + {\left( {\dfrac{5}{2}} \right)^2}$ becomes,
$ \Rightarrow {x^2} - 2x\left( {\dfrac{5}{2}} \right) + {\left( {\dfrac{5}{2}} \right)^2} = {\left( {x - \dfrac{5}{2}} \right)^2}$
Hence substituting this expression in the equation (2), we get,
$ \Rightarrow {\left( {x - \dfrac{5}{2}} \right)^2} = \dfrac{{61}}{4}$
Taking square root on both sides of the equation we get,
$ \Rightarrow x - \dfrac{5}{2} = \pm \sqrt {\dfrac{{61}}{4}} $
This can be written as,
$ \Rightarrow x - \dfrac{5}{2} = \pm \dfrac{{\sqrt {61} }}{{\sqrt 4 }}$
We know that $\sqrt 4 = 2$. Hence we get,
$ \Rightarrow x - \dfrac{5}{2} = \pm \dfrac{{\sqrt {61} }}{2}$
Adding $\dfrac{5}{2}$ on both sides, we get,
$ \Rightarrow x - \dfrac{5}{2} + \dfrac{5}{2} = \dfrac{5}{2} \pm \dfrac{{\sqrt {61} }}{2}$
$ \Rightarrow x + 0 = \dfrac{5}{2} \pm \dfrac{{\sqrt {61} }}{2}$
$ \Rightarrow x = \dfrac{5}{2} \pm \dfrac{{\sqrt {61} }}{2}$
This can be written as,
$ \Rightarrow x = \dfrac{{5 \pm \sqrt {61} }}{2}$

Hence the solution of the equation ${x^2} - 5x = 9$ by completing the square is given by $x = \dfrac{{5 \pm \sqrt {61} }}{2}$.

Note: Here many students solve the equation by using the quadratic formula given by
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. Since it is mentioned already in the question, they have to solve this kind of question by completing the square.
Method for completing the square is explained below.
To solve the quadratic equation of the form $a{x^2} + bx + c = 0$ by completing the square:
(1) If a, the leading coefficient is not equal to 1, then divide the whole equation by a.
(2) Transform the quadratic equation so that the constant term c is alone on the right hand side.
(3) Add the square of half the coefficient of x term, ${\left( {\dfrac{b}{{2a}}} \right)^2}$to both sides of the equation.
(4) Factor the left hand side as the square of the binomial.
(5) At last solve for the variable x.