Math Doubts

Solve the quadratic equation $2x^2 -7x + 3 = 0$ by Completing the Square

$2x^2 -7x + 3 = 0$ is an example for quadratic equation. Let’s solve this quadratic equation by completing the square method.

Move constant to Right side

Make some basic adjustment to transform the quadratic equation as a square of a binomial.

$\implies$ $2x^2 -7x$ $=$ $-3$

Divide equation by the coefficient of x2 term

In this example, the coefficient of $x^2$ is $2$. Divide both sides of the equation by the coefficient of $x^2$ term.

$\implies$ $\dfrac{2x^2 -7x}{2}$ $=$ $-\dfrac{3}{2}$

$\implies$ $\dfrac{2x^2}{2} -\dfrac{7x}{2}$ $=$ $-\dfrac{3}{2}$

$\implies$ $\Bigg(\dfrac{2}{2}\Bigg)x^2 -\Bigg(\dfrac{7}{2}\Bigg)x$ $=$ $-\dfrac{3}{2}$

$\implies$ $\require{cancel} \Bigg(\dfrac{\cancel{2}}{\cancel{2}}\Bigg)x^2 -\Bigg(\dfrac{7}{2}\Bigg)x$ $=$ $-\dfrac{3}{2}$

$\implies$ $x^2 -\Bigg(\dfrac{7}{2}\Bigg)x$ $=$ $-\dfrac{3}{2}$

Convert Algebraic expression as square of Binomial

The algebraic expression in this equation can be transformed as square of sum or difference of two terms. For example ${(a+b)}^2$ or ${(a-b)}^2$.

$\implies x^2 -x\Bigg(\dfrac{7}{2}\Bigg) = -\dfrac{3}{2}$

The second term should have $2$ as a factor but there is no $2$ as factor in the second term. Therefore, include $2$ as a factor in an acceptable manner in the second term of the algebraic expression.

$\implies$ $x^2 -\Bigg(\dfrac{2}{2}\Bigg)x\Bigg(\dfrac{7}{2}\Bigg)$ $\,=\,$ $-\dfrac{3}{2}$

$\implies$ $x^2 -2 \times \dfrac{1}{2} \times x\Bigg(\dfrac{7}{2}\Bigg)$ $\,=\,$ $-\dfrac{3}{2}$

$\implies$ $x^2 -2x\Bigg(\dfrac{7 \times 1}{2 \times 2}\Bigg)$ $\,=\,$ $-\dfrac{3}{2}$

$\implies$ $x^2 -2x\Bigg(\dfrac{7}{4}\Bigg)$ $\,=\,$ $-\dfrac{3}{2}$

In the second term of the algebraic expression, the second factor $x$ represents the first term and third factor represents third factor, if this expression is compared with the expansion of square of difference of two terms but there is no term in this expression. So, add and subtract the square of this factor in the expression.

$\implies$ $x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2 -{\Bigg(\dfrac{7}{4}\Bigg)}^2$ $\,=\,$ $-\dfrac{3}{2}$

$\implies$ $x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2 -\dfrac{49}{16}$ $\,=\,$ $-\dfrac{3}{2}$

$\implies$ $x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2$ $\,=\,$ $\dfrac{49}{16} -\dfrac{3}{2}$

$\implies$ $x^2 -2x\Bigg(\dfrac{7}{4}\Bigg) + {\Bigg(\dfrac{7}{4}\Bigg)}^2$ $\,=\,$ $\dfrac{49 -24}{16}$

The algebraic expression represents the expansion of square of difference of the two terms. So, it can be simplified mathematically as follows.

$\implies$ ${\Bigg(x -\dfrac{7}{4}\Bigg)}^2$ $\,=\,$ $\dfrac{25}{16}$

Solve the Quadratic Equation

Now, find the values of $x$ for solving the quadratic equation.

$\implies \Bigg(x -\dfrac{7}{4}\Bigg) = \pm \sqrt{\dfrac{25}{16}}$

$\implies x -\dfrac{7}{4} = \pm \dfrac{5}{4}$

$\implies x = \dfrac{7}{4} \pm \dfrac{5}{4}$

$\implies x = \dfrac{7 \pm 5}{4}$

$\implies x = \dfrac{7 + 5}{4}$ and $x = \dfrac{7 -5}{4}$

$\implies x = \dfrac{12}{4}$ and $x = \dfrac{2}{4}$

$\,\,\, \therefore \,\,\,\,\,\, x = 3$ and $x = \dfrac{1}{2}$

Therefore, the roots of quadratic equation $2x^2 -7x + 3 = 0$ by using completing the square method are $3$ and $\dfrac{1}{2}$.

Math Doubts

A best free mathematics education website for students, teachers and researchers.

Maths Topics

Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.

Maths Problems

Learn how to solve the math problems in different methods with understandable steps and worksheets on every concept for your practice.

Learn solutions

Subscribe us

You can get the latest updates from us by following to our official page of Math Doubts in one of your favourite social media sites.

Copyright © 2012 - 2022 Math Doubts, All Rights Reserved