Question

Finding roots of bi quadratic equation by factorizing

Answer 1

When bi quadratic eqn r given we usually take \(x^2\) as t & then solve the eqn but when the value of x^2 comes out to be an imaginary number then it is quite complex . However we use demoiver's theorem here & gt the roots.

we can find roots by factorizing the bi quadratic eqn into two quadratic eqn. For example:-

\[2x^4+2x^2+3=0\]\[x^4+x^2+\frac{3}{2}=0\]\[\left( x^4+\frac{3}{2}+2.\sqrt{\frac{3}{2}}x^2 \right)-2\sqrt{\frac{3}{2}}x^2+x^2=0\]\[\left( x^2+\sqrt{\frac{3}{2}} \right)^2-x^2(\sqrt{6}-1)\]\[\left( x^2+\sqrt{\frac{3}{2}} \right)^2-[x \sqrt{(\sqrt{6}-1)}]^2=0\]\[\left( x^2+\sqrt{\frac{3}{2}}+x \sqrt{\sqrt{6}-1} \right)\left( x^2+\sqrt{\frac{3}{2}}-x \sqrt{\sqrt{6}-1} \right)=0\]Now u have gt two quadratic eqn now u can find all four roots of biquadratic eqn :)

Answer 2

Nice!

Answer 3

thanx sir :)

Answer 4

@robtobey @kropot72 @Kainui @mayankdevnani

Answer 5

@Astrophysics

Answer 6

@Directrix @mathmale

Answer 7

@Michele_Laino

Answer 8

Is the question asking you to find the roots of:

2x^4 + 2x^2 + 3 = 0 ? I am not clear on the directions. @jiteshmeghwal9

Answer 9

yes @Directrix

Answer 10

@TheViper @the_ocean_girl @Preetha @satellite73 @amistre64 @UnkleRhaukus @Hero @darthsid

Answer 11

Thank you @jiteshmeghwal9 That was so informative.
Thanks to let me know about this.

Answer 12

:)