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WBJEE · Maths · Quadratic Equation

Let \(p, q\) be real numbers. If \(\alpha\) is the root of \(x^{2}+3 p^{2} x+5 q^{2}=0, \quad \beta \quad\) is \(\quad\) a \(\quad\) root \(\quad\) of \(x^{2}+9 p^{2} x+15 q^{2}=0\) and \(0 < \alpha < \beta,\) then the equation \(x^{2}+6 p^{2} x+10 q^{2}=0\) has a root \(\gamma\) that always satisfies

  1. A \(\gamma=\frac{\alpha}{4}+\beta\)
  2. B \(\beta < \gamma\)
  3. C \(\gamma=\frac{\alpha}{2}+\beta\)
  4. D \(\alpha < \gamma < \beta\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\alpha < \gamma < \beta\)

Step-by-step Solution

Detailed explanation

Since, \(\alpha\) is a root of \(\therefore\) \[ \begin{array}{l} x^{2}+3 p^{2} x+5 q^{2}=0 \\ \alpha^{2}+3 p^{2} \alpha+5 q^{2}=0 \end{array} \] and \(\beta\) is a root of \(\therefore\)…