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Question

Let α, β are the roots of the equation ax2+bx+c=0,(b0) and α4, β4 are the roots of the equation lx2+mx+n=0. If α, β are real and distinct, then the roots of the equation a2lx24aclx+2c2l+a2m=0 are
  1. always real
  2. opposite in sign
  3. same in sign
  4. always imaginary


Solution

The correct options are
A always real
B opposite in sign
α+β=ba, αβ=ca
α4+β4=ml, α4β4=nl
Now,
α4+β4=ml
(α2+β2)22α2β2=ml
((α+β)22αβ)22α2β2=ml
(b2a22ca)22c2a2=ml
(b2a2)24(b2a2)(ca)+2(c2a2)+ml=0
a2(b2a2)24ac(b2a2)+2c2+a2ml=0
a2l(b2a2)24acl(b2a2)+2c2l+a2m=0
Therefore,
b2a2 is a root of the equation
a2lx24aclx+2c2l+a2m=0

Also, b2a2>0
Thus, the equation a2lx24aclx+2c2l+a2m=0 has a positive real root.
Let γ be the other root of this equation.
Then, γ+b2a2=4acla2l
γ=4cab2a2=4acb2a2
γ<0
Hence, roots are always real and opposite in sign.

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