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Question

Let p and qbe real numbers. If α is the root of x2+3p2x+5q2=0,β is the root of x2+9p2x+15q2=0and 0<α<β, then the equation x2+6p2x+10q2=0 has root γ that always satisfies.


A

ɣ=ɑ4+β

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B

β<γ

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C

γ=ɑ2+β

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D

α<β<γ

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Solution

The correct option is D

α<β<γ


Explanation for the correct option.

Finding root γ always satisfies

Given: α is the root of x2+3p2x+5q2=0,β is the root of x2+9p2x+15q2=0 on putting the values we get,

x2+3p2x+5q2=0,α2+3p2α+5q2=0...(i)x2+9p2x+15q2=0

β2+9p2β+15q2=0...(ii)

Now, let

f(x)=x2+6p2x+10q2f(α)=α2+6p2α+10q2=(α2+3p2α+5q2)+3p2α+5q2=0+3p2α+5q2f(α)>0

and again,

f(β)=β2+6p2β+10q2=(β2+9p2β+15q2)(3p2β+5q2)=0(3p2β+5q2)f(β)<0

Thus, f(x) is a polynomial such that f(α)>0 and f(β)<0.
Therefore, there exists γ satisfying α<γ<βsuch that f(γ)=0.

Therefore, the correct answer is option (D).


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