TS EAMCET · Maths · Quadratic Equation
If one of the roots of the equation \(6 x^3-25 x^2+2 x+8=0\) is an integer and \(\alpha>0\), \(\beta < 0\) are the other two roots, then \(\frac{4}{\alpha}+\frac{1}{\beta}=\)
- A \(0\)
- B \(1\)
- C \(-2\)
- D \(4\)
Answer & Solution
Correct Answer
(D) \(4\)
Step-by-step Solution
Detailed explanation
\( 6(4)^3 - 25(4)^2 + 2(4) + 8 = 384 - 400 + 8 + 8 = 0 \) \( (6x^3-25x^2+2x+8) \div (x-4) = 6x^2-x-2 \) \( 6x^2-x-2=0 \Rightarrow x = \frac{-(-1) \pm \sqrt{(-1)^2-4(6)(-2)}}{2(6)} \) \( x = \frac{1 \pm \sqrt{1+48}}{12} = \frac{1 \pm 7}{12} \)…
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