Math Doubts

Solve $2x^2+x-1$ $=$ $0$ by Quadratic formula

A quadratic equation $2x^2+x-1 = 0$ is given in this problem and its roots (or zeros) should be calculated by using the quadratic formula. So, let’s learn how to find the zeros of the quadratic equation $2x^2+x-1 = 0$.

Compare the equation with standard form

Let us compare the given quadratic equation $2x^2+x-1 \,=\, 0$ with standard form of the quadratic equation $ax^2+bx+c \,=\,0$.

  1. $a = 2$
  2. $b = 1$
  3. $c = -1$

Substitute the values in Quadratic formula

It is time to substitute the values of them in the quadratic formula.

$x \,=\, \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\implies$ $x \,=\, \dfrac{-1\pm \sqrt{1^2-4 \times 2 \times (-1)}}{2 \times 2}$

Find the roots of equation by simplification

Now, simplify the quadratic formula to find the zeros or the given quadratic equation $2x^2+x-1$ $=$ $0$.

$\implies$ $x \,=\, \dfrac{-1\pm \sqrt{1+8}}{4}$

$\implies$ $x \,=\, \dfrac{-1\pm \sqrt{9}}{4}$

$\implies$ $x \,=\, \dfrac{-1\pm 3}{4}$

$\implies$ $x \,=\, \dfrac{-1+3}{4}$ or $x \,=\, \dfrac{-1-3}{4}$

$\implies$ $x \,=\, \dfrac{2}{4}$ or $x \,=\, \dfrac{-4}{4}$

$\implies$ $x \,=\, \dfrac{2}{4}$ or $x \,=\, -\dfrac{4}{4}$

$\implies$ $x \,=\, \dfrac{\cancel{2}}{\cancel{4}}$ or $x \,=\, -\dfrac{\cancel{4}}{\cancel{4}}$

$\,\,\,\therefore\,\,\,\,\,\,$ $x \,=\, \dfrac{1}{2}$ or $x \,=\, -1$

The solution set of the given quadratic equation $2x^2+x-1$ $=$ $0$ is $\bigg\{-1, \dfrac{1}{2}\bigg\}$

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