Question

If the roots of the equation $1,\frac{1}{2},\frac{1}{4}$be $4x^{3} + 16x^{2} - 9x - 36 = 0$and $- 2,\frac{2}{3}, - \frac{2}{3}$, then the roots of the equation $- 3,\frac{3}{2}, - \frac{3}{2}$ are.

• A. $- 4,\frac{3}{2}, - \frac{3}{2}$
• B. $2x^{5} - 14x^{4} + 31x^{3} - 64x^{2} + 19x + 130 = 0$
• C. $x^{3} - 3x + 2 = 0$
• D. None of these

Answer 1

Answer: (C)

$x^{3} - 3x + 2 = 0$

Answer 2

$x^{2} - (1 + n^{2})x + \frac{1}{2}(1 + n^{2} + n^{4}) = 0$are roots of $\alpha^{2} + \beta^{2}$

⇒ $2n$and $n^{3}$

Let the roots of $n^{2}$be $2n^{2}$, then

$x^{2} - 30x + p = 0$and $- 216$

but $x^{2} - 3x + 1 = 0$⇒$- 1$

Hence $\frac{1}{6}$ and $x^{2} + x - 6 = 0$.