Solve the equations:
$6 x^{4} - 29 x^{3} + 40 x^{2} - 7 x - 12 = 0$ , the product of two of the roots being $2$ .

Open in App

Solution

Given equation, $6 x^{4} - 29 x^{3} + 40 x^{2} - 7 x - 12 = 0$ ----------------(1)

Let the roots of the equation be $α, β, γ$ and $δ$

Product of the roots $= α β γ δ = - 2$

We are given that $α β = 2 ⟹ γ δ = - 1$

$∴$ the given equation can be written as $(A x^{2} + B x - A) (x^{2} + C x + 2) = 0$

$⟹ A^{4} + (B + A C) x^{3} + (A + B C) x^{2} + (2 B - A C) x - 2 A = 0$ ------------------(2)

Comparing the coefficients of (1) and (2), we get $A = 6, B = - 12$ and $C = - 3$

$∴$ the give equation can be written as $(6 x^{2} - 12 x - 6) (x^{2} - 3 x + 2) = 0$

Solving the two quadratic equation, we get the roots of the equation as

$x = 1, 2, 1 + \sqrt{2}$ and $1 - \sqrt{2}$

Suggest Corrections

Similar questions

(6 x^{4} - 29 x^{3} + 40 x^{2} - 12) \div (x - 3)

6 x^{4} - 13 x^{3} - 35 x^{2} - x + 3 = 0

2 - \sqrt{3}

4 x^{3} + 16 x^{2} - 9 x - 36 = 0

x^{3}

^{2}

6 x^{4} - 13 x^{3} - 35 x^{2} - x + 3 = 0

2 - \sqrt{3}

Join BYJU'S Learning Program