If the quadratic equation 3x2-4x-5-0 has roots - and - then the equation whose roots are 1- and 1- is

3x^2 + 4x + 5 = 0 #10; #10; α #160; + β #160; = 43 αβ #160; #10;= 53 #10; #10; 1α + 1β = α #160; + β #10;αβ = 4353 = ...

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Standard X

Mathematics

Roots of Quadratic Equation

If the quadra...

Question

If the quadratic equation $3 x^{2} + 4 x + 5 = 0$ has roots $α$ and $β$ then the equation whose roots are $\frac{1}{α}$ and $\frac{1}{β}$ is

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Solution

$3 x^{2} + 4 x + 5 = 0$

$α + β = \frac{- 4}{3} α β = \frac{5}{3}$

$\frac{1}{α} + \frac{1}{β} = \frac{α + β}{α β} = \frac{\frac{- 4}{3}}{\frac{5}{3}} = \frac{- 4}{5}$

$\frac{1}{α} \times \frac{1}{β} = \frac{1}{α β} = \frac{1}{(\frac{5}{3})} = \frac{3}{5}$

Required quadrant equation

$= x^{2} + (\frac{1}{α} + \frac{1}{β}) x + \frac{1}{α} + \frac{1}{β}$

$= x^{2} + (\frac{- 4}{5}) x + \frac{3}{5}$

$= x^{2} - \frac{4}{5} x + \frac{3}{5}$

$= 5 (5 x^{2} - 4 x + 3)$

Hence required equation is $5 x^{2} - 4 x + 3$

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If the quadratic equation 3x2+4x+5=0 has roots α and β then the equation whose roots are 1α and 1β is

3x2+4x+5=0 α+β=−43αβ=53 1α+1β=α+βαβ=−4353=−45 1α×1β=1αβ=1(53)=35 Required quadrant equation =x2+(1α+1β)x+1α+1β =x2+(−45)x+35 =x2−45x+35 =5(5x2−4x+3) Hence required equation is 5x2−4x+3

If the quadratic equation $3 x^{2} + 4 x + 5 = 0$ has roots $α$ and $β$ then the equation whose roots are $\frac{1}{α}$ and $\frac{1}{β}$ is