If - - are the roots of the equation 3 x 2-5 x -7-0- then the equation whose roots are 1-3 -5- 1-3 -5 isA- 147 x2-35 x-5-0B- 147 x 2-35 x -14-0- 21 x2-5 x-3-0D- 21 x2-5 x-1-0

The correct option is D 21x^2 +5x 1 = 0Let f(x) = 3x^2 +5x 7 = 0 has roots α, β So, 3α^2 +5α 7 = 0 ⇒ nbsp;3α+5 = 7α Now, having roots as nbsp;13α+5, 13β ...

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Byju's Answer

Standard XII

Mathematics

Relation between Roots and Coefficients for Quadratic

If α, β are t...

Question

If $α, β$ are the roots of the equation $3 x^{2} + 5 x - 7 = 0$ , then the equation whose roots are $\frac{1}{3 α + 5}, \frac{1}{3 β + 5}$ is

147 x^{2} + 35 x - 5 = 0

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147 x^{2} + 35 x - 14 = 0

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21 x^{2} + 5 x - 1 = 0

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21 x^{2} + 5 x - 3 = 0

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Solution

The correct option is C $21 x^{2} + 5 x - 1 = 0$
Let $f (x) = 3 x^{2} + 5 x - 7 = 0$ has roots $α, β$
So,
$3 α^{2} + 5 α - 7 = 0 \Rightarrow 3 α + 5 = \frac{7}{α}$
Now, having roots as $\frac{1}{3 α + 5}, \frac{1}{3 β + 5}$ is same as having roots $\frac{α}{7}, \frac{β}{7}$
The required equation is,
$f (7 x) = 0 \Rightarrow 3 (7 x)^{2} + 5 (7 x) - 7 = 0 ∴ 21 x^{2} + 5 x - 1 = 0$

Alternate Solution:
$\frac{1}{3 x + 5} = y$
$\Rightarrow x = \frac{1 - 5 y}{3 y}$
Therefore, equation whose roots are $\frac{1}{3 α + 5}, \frac{1}{3 β + 5}$ is
$f (\frac{1 - 5 x}{3 x}) = 0$
$\Rightarrow 3 {(\frac{1 - 5 x}{3 x})}^{2} + 5 (\frac{1 - 5 x}{3 x}) - 7 = 0$
$\Rightarrow 21 x^{2} + 5 x - 1 = 0$

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If α,β are the roots of the equation 3x2+5x−7=0, then the equation whose roots are 13α+5,13β+5 is

If $α, β$ are the roots of the equation $3 x^{2} + 5 x - 7 = 0$ , then the equation whose roots are $\frac{1}{3 α + 5}, \frac{1}{3 β + 5}$ is