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Algebra

Blazing goIITian

Joined:	26 Dec 2007
Post:	453

30 Dec 2007 10:19:36 IST

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429

A DOUBT IN QUADRATIC.

None

how to solve biquadratic equations? (3-x)⁴+(5-x)⁴=16. find the no of real roots.

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A K

Blazing goIITian

Joined: 13 Mar 2007

Posts: 700

30 Dec 2007 10:37:59 IST

Like 1 people liked this

try taking 3-x as Y^1/2
and 5-x as Y^1/2+2
then try to solve
u will get an equation as y^3/2 +10 y^1/2+8+2y=0
this is a cubic eqn in y^1/2
now take y^1/2 as p
so u get p^3+10p+8+2p^2
differentiating the above equation we see that the resulting quadriatic eqn has no real root
so the ans to the above question is 1
two imaginary roots have already been found out
the third root cannot be imaginary bcoz they always occur in pairs
so the third root is real
so y^1/2
has 1 real root and so has x

A K

Blazing goIITian

Joined: 13 Mar 2007

Posts: 700

30 Dec 2007 10:41:37 IST

Like 1 people liked this

sorry sorry made a mistake

at one step u have to cancel out y^1/2 terms

so
ans is 2????

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