$2x^2+x-4 = 0$ is a quadratic equation and solution of this equation has to be calculated by the completing the square method in this problem.
The number $-4$ is a constant in the given quadratic equation and shift it to right-hand side of the quadratic equation.
$2x^2+x-4 = 0$
$\implies$ $2x^2+x = 4$
The literal coefficient of $x^2$ in the first term is $2$ and divide both sides of this mathematical equation by this number.
$\implies$ $\dfrac{2x^2+x}{2}$ $\,=\,$ $\dfrac{4}{2}$
$\implies$ $\dfrac{2x^2}{2}+\dfrac{x}{2}$ $\,=\,$ $\dfrac{4}{2}$
$\implies$ $\require{cancel} \dfrac{\cancel{2}x^2}{\cancel{2}}+\dfrac{x}{2}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{4}}{\cancel{2}}$
$\implies$ $x^2+\dfrac{x}{2}$ $\,=\,$ $2$
Observe both sides of the mathematical equation. One side of the equation is in algebraic form and other side is a constant. The signs of both terms in the algebraic expression are positive. So, the algebraic expression can be converted in square form as per square of sum of two terms formula.
$\implies$ $x^2+1 \times \dfrac{x}{2}$ $\,=\,$ $2$
$\implies$ $x^2+\dfrac{2}{2} \times \dfrac{x}{2}$ $\,=\,$ $2$
$\implies$ $x^2+2 \times \dfrac{1}{2} \times x \times \dfrac{1}{2}$ $\,=\,$ $2$
$\implies$ $x^2+2 \times x \times \dfrac{1}{2} \times \dfrac{1}{2}$ $\,=\,$ $2$
$\implies$ $x^2+2 \times x \times \dfrac{1 \times 1}{2 \times 2}$ $\,=\,$ $2$
$\implies$ $x^2+2 \times x \times \dfrac{1}{4}$ $\,=\,$ $2$
$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)$ $\,=\,$ $2$
The algebraic expression is almost appeared same as the expansion of square of sum of two terms formula but a term is missed. So, add and subtract that term for expressing it in the square form completely.
$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2-{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $2$
$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $2+{\Big(\dfrac{1}{4}\Big)}^2$
$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $2+\dfrac{1^2}{4^2}$
$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $2+\dfrac{1}{16}$
$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $\dfrac{2 \times 16 + 1}{16}$
$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $\dfrac{32+1}{16}$
$\implies$ $x^2+2x\Big(\dfrac{1}{4}\Big)+{\Big(\dfrac{1}{4}\Big)}^2$ $\,=\,$ $\dfrac{33}{16}$
$\implies$ ${\Big(x+\dfrac{1}{4}\Big)}^2$ $\,=\,$ $\dfrac{33}{16}$
$\implies$ $x+\dfrac{1}{4}$ $\,=\,$ $\pm \sqrt{\dfrac{33}{16}}$
$\implies$ $x+\dfrac{1}{4}$ $\,=\,$ $\pm \dfrac{\sqrt{33}}{4}$
$\implies$ $x$ $\,=\,$ $-\dfrac{1}{4} \pm \dfrac{\sqrt{33}}{4}$
$\implies$ $x$ $\,=\,$ $\dfrac{-1 \pm \sqrt{33}}{4}$
$\,\,\, \therefore \,\,\,\,\,\,$ $x \,=\, \dfrac{-1+\sqrt{33}}{4}$ and $x \,=\, \dfrac{-1-\sqrt{33}}{4}$
A best free mathematics education website for students, teachers and researchers.
Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.
Learn how to solve the maths problems in different methods with understandable steps.
Copyright © 2012 - 2022 Math Doubts, All Rights Reserved