Question

How do you solve the quadratic equation by completing the square: $2{x^2} - 7x = 2$?

Answer 1

The given equation is quadratic equation of the form $a{x^2} + bx + c$ , in which x is an unknown term and as given here, we need to solve for x and completing the square method is one of the methods to find the roots of the given quadratic equation. A polynomial equation with degree equal to two is known as a quadratic equation.

Complete step by step answer:
Given,
$2{x^2} - 7x = 2$
Divide both sides of the equation by 2 as:
$\dfrac{{2{x^2}}}{2} - \dfrac{7}{2}x = \dfrac{2}{2}$
$\Rightarrow {x^2} - \dfrac{7}{2}x = 1$ …………. 1
To apply complete the square method means to force a perfect square trinomial of the equation in the form: ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$.
Hence, divide the coefficient of the x term by 2 in equation 1, square the result and add to both sides of the equation as we have equation 1 as:
${x^2} - \dfrac{7}{2}x = 1$
$\Rightarrow - \dfrac{{\dfrac{7}{2}}}{2} = \left( { - \dfrac{7}{2}} \right) \cdot \dfrac{1}{2}$
As, we have divided the coefficient of the x term by 2, hence we get:
$= - \dfrac{7}{4}$ ……………………. 2
Now, as mentioned we need to square the result and add to both sides of the equation 2 as:
$\Rightarrow {\left( { - \dfrac{7}{4}} \right)^2} = \dfrac{{49}}{{16}}$
Now, add to both sides of the equation by obtained result as $\dfrac{{49}}{{16}}$ in equation 1:
As we have equation 1 as: ${x^2} - \dfrac{7}{2}x = 1$
$\Rightarrow {x^2} - \dfrac{7}{2}x + \dfrac{{49}}{{16}} = 1 + \dfrac{{49}}{{16}}$
The common denominator for 1 and $\dfrac{{49}}{{16}}$ is 16. Multiply 1 time $\dfrac{{16}}{{16}}$, then add the two fractions as:
${x^2} - \dfrac{7}{2}x + \dfrac{{49}}{{16}} = \dfrac{{16}}{{16}} + \dfrac{{49}}{{16}}$
As the denominator consists of common term 16, hence combine both the terms as:
${x^2} - \dfrac{7}{2}x + \dfrac{{49}}{{16}} = \dfrac{{16 + 49}}{{16}}$
Hence, adding both the terms, we get:
$\Rightarrow {x^2} - \dfrac{7}{2}x + \dfrac{{49}}{{16}} = \dfrac{{65}}{{16}}$ ………………. 3
We now have a perfect square trinomial as in equation 3, where $a = x$and $b = \dfrac{7}{4}$.
${\left( {x - \dfrac{7}{4}} \right)^2} = \dfrac{{65}}{{15}}$
Now, take square root on both the sides:
$\left( {x - \dfrac{7}{4}} \right) = \pm \sqrt {\dfrac{{65}}{{16}}}$
We know that ${4^2} = 16$i.e., $\sqrt {16} = 4$, hence we get:
$\Rightarrow \left( {x - \dfrac{7}{4}} \right) = \pm \dfrac{{\sqrt {65} }}{4}$
Solve for x we get:
$x = \dfrac{7}{4} \pm \dfrac{{\sqrt {65} }}{4}$
Hence, after solving the quadratic equation by completing the square we get the value of x as:
$x = \dfrac{7}{4} \pm \dfrac{{\sqrt {65} }}{4}$ or $x = \dfrac{{7 \pm \sqrt {65} }}{4}$`

Note: We have used here completing the square technique in which it is a technique which can be used to find maximum or minimum values of quadratic functions. We can also use this technique to change or simplify the form of algebraic expressions. Hence, it is used for solving quadratic equations. As we have used here of the form ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$.