Question

Q49
Q49. If a and are the roots of the equation 2x² – 3x-6= 0.
The equation whose roots are and is1/a and 1/b is

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Answer 1

$2 {x}^{2} - 3x - 6 = 0$

sum of the roots , a+b = 3/2.

product of the roots, ab = -3.

$\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} = \frac{ \frac{3}{2} }{ - 3} = \frac{ - 1}{2}$

$\frac{1}{a} \times \frac{1}{b} = \frac{1}{ab} = - \frac{1}{3}$

equation =

${x}^{2} - ( \frac{1}{a} + \frac{1}{b} )x + ( \frac{1}{a} \times \frac{1}{b} ) = 0$

${x}^{2} - ( - \frac{1}{2} )x + ( - \frac{1}{3} ) = 0$

${x}^{2} + \frac{1}{2} x - \frac{1}{3} = 0$

$\frac{6 {x}^{2} + 3x - 2}{6} = 0$

Therefore, equation is

$6 {x}^{2} + 3x - 2 = 0$

Answer 2

$2 {x}^{2} - 3x - 6 = 0$

sum of the roots = a + b = 3/2

product of the roots = ab = -3

$\frac{1}{a} + \frac{1}{b} = \frac{a + b}{ab} = \frac{ \frac{3}{2} }{ - 3} = - \frac{1}{2}$

$\frac{1}{a} \times \frac{1}{b} = \frac{1}{ab} = - \frac{1}{3}$

The equation with the roots 1/a , 1/b is -

${x}^{2} - ( \frac{1}{a} + \frac{1}{b} )x + ( \frac{1}{a} \times \frac{1}{b} ) = 0$

${x}^{2} + \frac{1}{2} x - \frac{1}{3} = 0$

$\frac{6 {x}^{2} + 3x - 2 }{6} = 0$

$6{x}^{2} + 3x - 2 = 0$

Therefore, the final equation is -

$6 {x}^{2} + 3x - 2 = 0$