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If x1 and x2 are the roots of the equation 2x2 + 3x – 9 = 0 then the equation which has the roots 1/x1 and 1/x2 is:
- A9x2 + 3x – 2 = 0
- B-9x2 – 3x – 2 = 0
- C9x2 – 3x – 2 = 0
- D9x2 – 3x + 2 = 0
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CCommunity
We know that, for a quadratic equation ax2 + bx + c,
Sum of its roots = -b/a and product of roots = c/a
For 2x2 + 3x – 9 = 0:
x1 + x2 = -3/2 and x1x2 = -9/2
For the equation whose roots are 1/x1 and 1/x2,
Sum of roots $(= \frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} = \frac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} = \frac{{ - 3}}{{ - 9}} = \frac{1}{3})$
Product of roots $(= \frac{1}{{{x_1}{x_2}}} = \; - \frac{2}{9})$
∴ the quadratic equation becomes: $({x^2} - \frac{1}{3}x - \frac{2}{9} = 9{x^2} - 3x - 2)$
Sum of its roots = -b/a and product of roots = c/a
For 2x2 + 3x – 9 = 0:
x1 + x2 = -3/2 and x1x2 = -9/2
For the equation whose roots are 1/x1 and 1/x2,
Sum of roots $(= \frac{1}{{{x_1}}} + \frac{1}{{{x_2}}} = \frac{{{x_1} + {x_2}}}{{{x_1}{x_2}}} = \frac{{ - 3}}{{ - 9}} = \frac{1}{3})$
Product of roots $(= \frac{1}{{{x_1}{x_2}}} = \; - \frac{2}{9})$
∴ the quadratic equation becomes: $({x^2} - \frac{1}{3}x - \frac{2}{9} = 9{x^2} - 3x - 2)$