Question

Statement I: lf $\alpha ,\beta$ are the roots of $x^2-ax+b=0$, then the equation whose roots are $\frac{\alpha +\beta }{\alpha },\frac{\alpha +\beta }{\beta }$ is $b{x}^2-a^{2}x+a^2=0$
Statement II: lf $\alpha ,\beta$ are the roots of $x^2-bx+c=0$ and $\alpha +h,\beta +h$ are the roots of $x^2+qx+r=0$, then $h=b-q$.
Which of the above statement(s) is(are) true.

• A. only I
• B. only II
• C. both I and II
• D. neither I and II

Answer 1

The correct option is A only I

Statement I: As $\alpha ,\beta$ are roots of $x^2-ax+b=0,$ then

$S_1=\alpha +\beta =a$ and $S_2=\alpha \beta =b$

Now $\frac{\alpha +\beta }{\alpha }+\frac{\alpha +\beta }{\beta }=\frac{\beta (\alpha +\beta )+\alpha (\alpha +\beta )}{\alpha \beta }=\frac{(\alpha +\beta )(\alpha +\beta )}{\alpha \beta }=\frac{a^2}{b}$

And $\frac{\alpha +\beta }{\alpha }\times \frac{\alpha +\beta }{\beta }=\frac{(\alpha +\beta )(\alpha +\beta )}{\alpha \beta }=\frac{a^2}{b}$

Then equation whose roots are $\frac{\alpha +\beta }{\alpha },\frac{\alpha +\beta }{\beta }$ is

$x^2-\left(\frac{a^2}{b}\right)x+\left(\frac{a^2}{b}\right)=0\Rightarrow b{x}^2-a^{2}x+a^2=0$

Statement II: AS $\alpha ,\beta$ are roots of $x^2-bx+c=0,$ then

$S_1=\alpha +\beta =b$ and $S_2=\alpha \beta =c$

And as $\alpha +h,\beta +h$ are roots of $x^2+qx+r=0,$ then

${S^{\prime }}_1=\alpha +h+\beta +h=-q\Rightarrow \alpha +\beta +2h=-q\Rightarrow h=\frac{-q-b}{2}$

And ${S^{\prime }}_2=(\alpha +h)(\beta +h)=r\Rightarrow \alpha \beta +(\alpha +\beta )h+h^2=r$

$\Rightarrow c+bh+h^2=r\Rightarrow h=\frac{-b\pm \sqrt{b^2-4(c-r)}}{2}$

Hence statement II is false and only statement I is true.