6x2-1-x2-25x-1-x-12-0

The correct option is A , , 2, 3Let x 1/x=z (x 1/x)^2=z^2 (x^2+1/x^2 2)=z^2 x^2+1/x^2=z^2+2 Now substituting in the equation: 6(z^2+2) 25z+12=0 6z^2+12 25z ...

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Quantitative Aptitude

Quadratic Equations

6x2+1/x2-25x-...

Question

$6 (x^{2} + \frac{1}{x^{2}}) - 25 (x - \frac{1}{x}) + 12 = 0$

, 2, 3

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, -2, -3

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, 2, 3

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Solution

The correct option is A , , 2, 3
Let $x - \frac{1}{x} = z$
$(x - \frac{1}{x})^{2} = z^{2}$
$(x^{2} + \frac{1}{x^{2}} - 2) = z^{2}$
$x^{2} + \frac{1}{x^{2}} = z^{2} + 2$
Now substituting in the equation:
$6 (z^{2} + 2) - 25 z + 12 = 0$
$6 z^{2} + 12 - 25 z + 12 = 0$
$6 z^{2} - 25 z + 24 = 0$
$6 z^{2} - 16 z - 9 z + 24 = 0$
$2 z (3 z - 8) - 3 (3 z - 8) = 0$
$(2 z - 3) (3 z - 8) = 0$
$z = \frac{3}{2}; z = \frac{8}{3}$
$x - \frac{1}{x} = \frac{3}{2}$
$x - \frac{1}{x} = \frac{8}{3}$
$2 x^{2} = 3 x - 2 = 0$
$3 x^{2} - 8 x - 3 = 0$
$2 x^{2} - 4 x + x - 2 = 0$
$3 x^{2} - 9 x + x - 3 = 0$
$2 x (x - 2) + 1 (x - 2) = 0$
$3 x (x - 3) + 1 (x - 3) = 0$
$(2 x + 1) (x - 2) = 0$
$(3 x + 1) (x - 3) = 0$
$x = - \frac{1}{2}, x = 2$
$x = - \frac{1}{3}, x = 3$
So,the roots are: $- \frac{1}{2}, 2, - \frac{1}{3}, 3$

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6(x2+1x2)−25(x−1x)+12=0

$6 (x^{2} + \frac{1}{x^{2}}) - 25 (x - \frac{1}{x}) + 12 = 0$