AP EAMCET · Maths · Quadratic Equation

If $α, β$ are the real roots of $x^{2} + p x + q = 0$ and $α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0,$ then the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$ has always

A two positive roots
B two negative roots
C one positive root and one negative root
D two real roots

Verified Solution

Answer & Solution

Correct Answer

(D) two real roots

Step-by-step Solution

Detailed explanation

It is given that, α, β are the roots of x2+px+q=0 α+β=-p and αβ=q Since, α4, β4 are roots of x2-rx+s=0 Therefore, α4+β4=r and α4β4=s Now, x2-4qx+2q2-r=0 D=(4q)2-42q2-r =16q2-8q2+4r =8q2+4r Here,…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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If α, β are the real roots of x2+px+q=0 and α4, β4 are the roots of x2-rx+s=0, then the equation x2-4qx+2q2-r=0 has always

Step-by-step Solution

If $α, β$ are the real roots of $x^{2} + p x + q = 0$ and $α^{4}, β^{4}$ are the roots of $x^{2} - r x + s = 0,$ then the equation $x^{2} - 4 q x + 2 q^{2} - r = 0$ has always