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STD 11 - 4.2 Quadratic equations and inequations — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 4.2 Quadratic equations and inequations4 Marks
MCQ
If $\alpha \ne \beta $ but ${\alpha ^2} = 5\alpha - 3$ and ${\beta ^2} = 5\beta - 3$, then the equation whose roots are $\alpha /\beta $ and $\beta /\alpha $ is
  • A
    $3{x^2} - 25x + 3 = 0$
  • B
    ${x^2} + 5x - 3 = 0$
  • C
    ${x^2} - 5x + 3 = 0$
  • ✓
    $3{x^2} - 19x + 3 = 0$
✓

Answer

Correct option: D.
$3{x^2} - 19x + 3 = 0$
d
(d) ${\alpha ^2} - 5\alpha + 3 = 0$ …..$(i)$

${\beta ^2} - 5\beta + 3 = 0$ …..$(ii)$

From $(i) -(ii),$

==> $({\alpha ^2} - {\beta ^2}) - 5\alpha + 5\beta = 0$

==> ${\alpha ^2} - {\beta ^2} = 5(\alpha - \beta )$$ \Rightarrow \alpha + \beta = 5$

From $(i) + (ii),$

==> $({\alpha ^2} + {\beta ^2}) - 5(\alpha + \beta ) + 6 = 0$

==> $({\alpha ^2} + {\beta ^2}) - 5.5 + 6 = 0$ ==> ${\alpha ^2} + {\beta ^2} = 19$

Then ${(\alpha + \beta )^2} = {\alpha ^2} + {\beta ^2} + 2\alpha \beta $

$ \Rightarrow 25 = 19 + 2\alpha \beta $ $ \Rightarrow \alpha \beta = 3$

then the equation, whose roots are $\frac{\alpha }{\beta }$ and $\frac{\beta }{\alpha }$, is

${x^2} - x\left( {\frac{\alpha }{\beta } + \frac{\beta }{\alpha }} \right) + \frac{\alpha }{\beta }.\frac{\beta }{\alpha } = 0$

==> ${x^2} - x\left( {\frac{{{\alpha ^2} + {\beta ^2}}}{{\alpha \beta }}} \right) + 1 = 0$

$ \Rightarrow {x^2} - x.\frac{{19}}{3} + 1 = 0$

==> $3{x^2} - 19x + 3 = 0$.

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