Name: Transformation of Roots of a Quadratic Equation
Uploaded: 2019-07-19
Duration: 11 min 45 s
Description: Here an equation and its roots are given, and we have to find new quadratic equations with new roots, which are simple multiples or factors of the original r...

Question

Solve the equation,

6 x^{4} - 29 x^{3} + 40 x^{2} - 7 x - 12 = 0;

the product of two of the roots being

2

.

Accepted Answer

Given equation is $6 x^{4} - 29 x^{3} + 40 x^{2} - 7 x - 12 = 0$ .

Let the four roots be $α, β, γ, δ$ and suppose $α β = 2$ .

Then $α + β + γ + δ = \frac{29}{6} . . . (1)$

$α β + α γ + α δ + β γ + β δ + γ δ = \frac{40}{6} . . . (2)$

$α β γ + α β δ + α γ δ + β γ δ = \frac{7}{6} . . . (3)$

and $α β γ δ = - 2 . . . (4)$

$∵ α β = 2 ∴ γ δ = - 1$ from $(4)$

From $(3)$ we have $α β (γ + δ) + γ δ (α + β) = \frac{7}{6}$ .

$\Rightarrow 2 (γ + δ) + (- 1) (α + β) = \frac{7}{6}$

$\Rightarrow 2 γ + 2 δ - α - β = \frac{7}{6}$

$\Rightarrow 2 γ + 2 δ + 2 α + 2 β - 3 α - 3 β = \frac{7}{6}$

$\Rightarrow 2 (α + β + γ + δ) - 3 (α + β) = \frac{7}{6}$

$\Rightarrow 2 (\frac{29}{6}) - 3 (α + β) = \frac{7}{6}$

$\Rightarrow 3 (α + β) = \frac{58}{6} - \frac{7}{6} = \frac{51}{6}$

$α + β = \frac{17}{6}$ and $γ + δ = 2$

Thus $\begin{matrix} α + β = \frac{17}{6} \\ α β = 2 \end{matrix}\}; \begin{matrix} γ + δ = 2 \\ γ δ = - 1 \end{matrix}\}$

If $α + β = \frac{17}{6}$ , then $β = \frac{17}{6} - α$

$α β = 2$

$\Rightarrow α (\frac{17}{6} - α) = 2$

$\Rightarrow 17 α - 6 α^{2} = 12$

$\Rightarrow 6 α^{2} - 17 α + 12 = 0$

$\Rightarrow 6 α^{2} - 8 α - 9 α + 12 = 0$

$\Rightarrow (3 α - 4) (2 α - 3) = 0$

$β = \frac{3}{2} or β = \frac{4}{3}$

Hence, $β = \frac{3}{2} or β = \frac{4}{3}$

If $γ + δ = 2$ , then $δ = 2 - γ$

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$\Rightarrow γ (2 - γ) = - 1$

$\Rightarrow 2 γ - γ^{2} = - 1$

$\Rightarrow γ^{2} - 2 γ - 1 = 0$

$\Rightarrow γ = \frac{2 \pm \sqrt{4 + 4}}{2}$

$\Rightarrow γ = 1 \pm \sqrt{2}$

$\Rightarrow γ = 1 + \sqrt{2} o r γ = 1 - \sqrt{2}$

Hence, $δ = 1 - \sqrt{2} or δ = 1 + \sqrt{2}$

$∴ α = \frac{4}{3}, β = \frac{3}{2}, γ = 1 + \sqrt{2}, δ = 1 - \sqrt{2}$

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

Important Questions on Theory of Equations

Learn and Prepare for any exam you want !

Solve the equation, 6x4-29x3+40x2-7x-12=0; the product of two of the roots being 2.

Important Questions on Theory of Equations

Learn and Prepare for any exam you want !

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