Solve quadratic equation for x 4x² …
CBSE, JEE, NEET, CUET
Question Bank, Mock Tests, Exam Papers
NCERT Solutions, Sample Papers, Notes, Videos
Posted by Nikita Oli 6 years, 2 months ago
- 2 answers
Related Questions
Posted by Parinith Gowda Ms 1 week, 1 day ago
- 0 answers
Posted by Vanshika Bhatnagar 1 year, 1 month ago
- 2 answers
Posted by Hari Anand 3 months ago
- 0 answers
Posted by Kanika . 1 month ago
- 1 answers
Posted by Sahil Sahil 1 year, 1 month ago
- 2 answers
Posted by Parinith Gowda Ms 1 week, 1 day ago
- 1 answers
Posted by Lakshay Kumar 9 months, 3 weeks ago
- 0 answers
myCBSEguide
Trusted by 1 Crore+ Students
Test Generator
Create papers online. It's FREE.
CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
Sia ? 6 years, 2 months ago
4x2-4a2x+(a4-b4)=0
a = 4, b = -4a{tex}^2{/tex}, c = {tex}a^4-b^4{/tex}
D = {tex}b^{2}-4 a c{/tex}
= {tex}\left(-4 a^{2}\right)^{2}-4 \times 4 \times\left(a^{4}-b^{4}\right){/tex}
{tex}=16 a^{4}-16 a^{4}+16 b^{4}{/tex}
D = 16 b4
Therefore,by quadratic formula,we have,
{tex}x=\frac{-b \pm \sqrt{D} }{ 2 a}{/tex}
{tex}=\frac{-\left(-4 a^{2}\right) \pm 4 b^{2}}{2 \times 4}{/tex}
{tex}=\frac{4 a^{2} \pm 4 b^{2}} {8}{/tex}
x = {tex}\frac{ a^{2}\pm b^2} {2}{/tex}
x = {tex}\frac{a^2+b^2}{2}{/tex} or x={tex}\frac{a^2-b^2}{2}{/tex}
{tex}{/tex} {tex}{/tex}Hence,the roots of given quadratic equation is ,{tex}\frac{a^2+b^2}{2} ,\frac{a^2-b^2}{2}{/tex}
1Thank You