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 Questions

The math problems with solutions to learn how to solve a problem.

Learn solutions

Math Worksheets

The math worksheets with answers for your practice with examples.

Practice now

Math Videos

The math videos tutorials with visual graphics to learn every concept.

Watch now

Subscribe us

Get the latest math updates from the Math Doubts by subscribing us.

Learn more

Math Doubts

A free math education service for students to learn every math concept easily, for teachers to teach mathematics understandably and for mathematicians to share their maths researching projects.

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved