Question

If $y^{3} + y + 2 = 0$ and $y^{3} - y^{2} - y - 2 = 0$ are the roots of the equation $y^{3} + 3y^{2} - y - 3 = 0$ and $y^{3} + 4y^{2} + 5y + 20 = 0$, then $2x^{3} - 3x^{2} + 6x + 1 = 0$

• A. –8
• B. –16
• C. 16
• D. 8

Answer 1

Answer: (B)

–16

Answer 2

$\mathbf{x}^{\mathbf{2}}\mathbf{+ px + q = 0}$ …..(i)

$2 + \sqrt{3}$ …..(ii)

and given $\sqrt{3}$ …..(iii)

Solving (i) and (iii), we get $- 2, - \sqrt{3}$

Substituting these values in (ii), we get $ax^{2} + bx + c = 0$.