TS EAMCET · Maths · Quadratic Equation
If both the roots of the equation \(x^2-4 a x+1-3 a+4 a^2=0\) exceed 1 , then \(a\) lies in the interval
- A \(\left(-\infty, \frac{7-\sqrt{17}}{8}\right)\)
- B \(\left(\frac{7+\sqrt{17}}{8}, \infty\right)\)
- C \(\left(\frac{7-\sqrt{17}}{8}, \frac{1}{2}\right)\)
- D \(\left(\frac{1}{2}, \frac{7+\sqrt{17}}{8}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{7+\sqrt{17}}{8}, \infty\right)\)
Step-by-step Solution
Detailed explanation
We have, \[ \begin{array}{ll} & x^2-4 a x+1-3 a+4 a^2=0 \\ \text { Let } & f(x)=x^2-4 a x+1-3 a+4 a^2 \end{array} \] Both roots of \(f(x)\) has greater than 1…
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