WBJEE · Maths · Quadratic Equation
If the equation \(x^{2}-ca+d=0\) has roots equal to the fourth powers of the roots of \(x^{2}+a x+b=0,\) where \(a^{2}>4 b,\) then the roots of \(x^{2}-4 b x+2 b^{2}-c=0\) will be
- A both real
- B both negative
- C both positive
- D one positive and one negative
Answer & Solution
Correct Answer
(D) one positive and one negative
Step-by-step Solution
Detailed explanation
Let \(\alpha\) and \(\beta\) are the roots of \(x^{2}+a x+b=0\) and the roots of \(x^{2}-cx+d=0\) are \(\alpha^{4}\) and \(\beta^{4}\) Now, \(\alpha+\beta=-a, \alpha \beta=b\) ...(i) and \(\alpha^{4}+\beta^{4}-c, \alpha^{4} \beta^{4}=d\) ...(ii) From Eqs. (i) and (ii),…
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