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Nature of Roots

if alpha and ...

Question

if alpha and beta are the roots of 2x²+ x+3 =0, then equation whose roots are 1-alpha/1+alpha and 1-beta/1+beta is

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Solution

$Dear Student, 2 x^{2} + x + 3 = 0 Here, roots are α and β, Now we know that; α + β = \frac{- b}{a} = \frac{- 1}{2} . . . . . . . . . (i) and, αβ = \frac{c}{a} = \frac{3}{2} . . . . . . . (ii) Now, The new quadratic equation has roots which are; \frac{1 - α}{1 + α} and \frac{1 - β}{1 + β} Now, Summation of roots; \frac{1 - α}{1 + α} + \frac{1 - β}{1 + β} = \frac{(1 + β) (1 - α) + (1 - β) (1 + α)}{(1 + α) (1 + β)} = \frac{1 + β - α - αβ + 1 + α - β - αβ}{1 + α + β + αβ} = \frac{2 (1 - αβ)}{1 + (α + β) + αβ} Now put value of (α + β) and αβ from (i) and (ii), we get; = \frac{2 (1 - \frac{3}{2})}{1 - \frac{1}{2} + \frac{3}{2}} = \frac{- 1}{2} Summation of roots of new quadratic equation = \frac{- 1}{2} Now, (\frac{1 - α}{1 + α}) \times (\frac{1 - β}{1 + β}) = \frac{1 - α - β + αβ}{1 + α + β + αβ} = \frac{1 - (α + β) + αβ}{1 + (α + β) + αβ} Now put value of (α + β) and αβ from (i) and (ii), we get; = \frac{1 + \frac{1}{2} + \frac{3}{2}}{1 - \frac{1}{2} + \frac{3}{2}} = \frac{3}{2} Product of roots of new quadratic equation = \frac{3}{2} We can write quadratic equation in the below mentioned form; x^{2} - (sum of roots) x + (product of roots) = 0 So, \Rightarrow x^{2} - (\frac{- 1}{2}) x + \frac{3}{2} = 0 \Rightarrow 2 x^{2} + x + 3 = 0 Which is identical as parent quadratic equation . Regards .$

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