![Trending Articles on Technical and Non Technical topics](/images/trending_categories.jpeg)
Data Structure
Networking
RDBMS
Operating System
Java
MS Excel
iOS
HTML
CSS
Android
Python
C Programming
C++
C#
MongoDB
MySQL
Javascript
PHP
Physics
Chemistry
Biology
Mathematics
English
Economics
Psychology
Social Studies
Fashion Studies
Legal Studies
- Selected Reading
- UPSC IAS Exams Notes
- Developer's Best Practices
- Questions and Answers
- Effective Resume Writing
- HR Interview Questions
- Computer Glossary
- Who is Who
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
$2x^2 + x - 4 = 0$
Given:
Given quadratic equation is $2x^2+x -4 = 0$.
To do:
We have to find the roots of the given quadratic equation.
Solution:
$2x^2+x - 4 = 0$
$2(x^2 + \frac{1}{2} x -\frac{4}{2}) = 0$
$x^2 + \frac{1}{2} x -2 = 0$
$x^2 + 2\times \frac{1}{2}\times \frac{1}{2} x = 2$
$x^2 + 2\times \frac{1}{4} x = 2$
Adding $(\frac{1}{4})^2$ on both sides completes the square. Therefore,
$x^2 + 2\times (\frac{1}{4}) x + (\frac{1}{4})^2 = 2+(\frac{1}{4})^2$
$(x+\frac{1}{4})^2=2+\frac{1}{16}$ (Since $(a+b)^2=a^2+2ab+b^2$)
$(x+\frac{1}{4})^2=\frac{1+2\times16}{16}$
$(x+\frac{1}{4})^2=\frac{1+32}{16}$
$(x+\frac{1}{4})^2=\frac{33}{16}$
$x+\frac{1}{4}=\pm \sqrt{\frac{33}{16}}$ (Taking square root on both sides)
$x=\sqrt{\frac{33}{16}}-\frac{1}{4}$ or $x=-\sqrt{\frac{33}{16}}-\frac{1}{4}$
$x=\frac{\sqrt{33}-1}{4}$ or $x=-(\frac{\sqrt{33}+1}{4})$
The values of $x$ are $\frac{\sqrt{33}-1}{4}$ and $-(\frac{\sqrt{33}+1}{4})$.