Question

If $\alpha ,\beta $ are the roots of the equation ${x^2} - 2x + 3 = 0$, then the equation whose roots are $\dfrac{1}{{{\alpha ^2}}}$ and \[\dfrac{1}{{{\beta ^2}}}\] is
A. ${x^2} + 2x + 1 = 0$
B. $9{x^2} + 2x + 1 = 0$
C. $9{x^2} - 2x + 1 = 0$
D. $9{x^2} + 2x - 1 = 0$

Answer 1

Hint: Find $\alpha + \beta $ and $\alpha \beta $ from the equation ${x^2} - 2x + 3 = 0$. Form an equation using $\dfrac{1}{{{\alpha ^2}}}$ and \[\dfrac{1}{{{\beta ^2}}}\] as the roots and expand the equation using the values of $\alpha + \beta $ and $\alpha \beta $.

Formula used:
${a^2} + {b^2} = {(a + b)^2} - 2ab$

Complete step by step Solution:
We know that given an equation $a{x^2} + bx + c = 0$, the sum of the roots is $\dfrac{{ - b}}{a}$ and the product of the roots is $\dfrac{c}{a}$.
Therefore, from the equation ${x^2} - 2x + 3 = 0$, $\alpha + \beta = \dfrac{{ - ( - 2)}}{1} = 2$ and $\alpha \beta = \dfrac{3}{1} = 3$.
An equation whose roots are $\delta ,\gamma $ can be written as ${x^2} - (\delta + \gamma )x + \delta \gamma = 0$.
Using this knowledge, we can write the equation whose roots are $\dfrac{1}{{{\alpha ^2}}}$ and \[\dfrac{1}{{{\beta ^2}}}\] as
${x^2} - (\dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}})x + \dfrac{1}{{{\alpha ^2}}}\dfrac{1}{{{\beta ^2}}} = 0$
${x^2} - (\dfrac{{{\alpha ^2} + {\beta ^2}}}{{{\alpha ^2}{\beta ^2}}})x + \dfrac{1}{{{\alpha ^2}}}\dfrac{1}{{{\beta ^2}}} = 0$
${x^2} - (\dfrac{{{{(\alpha + \beta )}^2} - 2\alpha \beta }}{{{{(\alpha \beta )}^2}}})x + \dfrac{1}{{{{(\alpha \beta )}^2}}} = 0$
Since \[\alpha + \beta = 2\], \[{(\alpha + \beta )^2} = 4\]. Similarly, since $\alpha \beta = 3$, \[{(\alpha \beta )^2} = 9\].
Substituting these values back to the equation, we get
${x^2} - (\dfrac{{4 - 6}}{9})x + \dfrac{1}{9} = 0$
\[{x^2} + \dfrac{2}{9}x + \dfrac{1}{9} = 0\]
$9{x^2} + 2x + 1 = 0$

Therefore, the correct option is (B).

Note:Given an equation in the standard form $a{x^2} + bx + c = 0$, the sum of the roots of the quadratic equation is given by $\dfrac{{ - b}}{a}$ and the product of the roots of the quadratic equation is given by $\dfrac{c}{a}$.