Question

A quadratic equation is chosen from the set of all quadratic equations which are unchanged by squaring their roots. The chance that the chosen equation has equal roots is

• A. $1/2$
• B. $1/3$
• C. $1/4$
• D. $2/3$

Answer 1

Answer: (B)

$1/3$

Given,the quadratice equation does not change even the roots are squared.
Let the roots of the quadratic equation be $p,q$.
After squaring, the new roots are $p^2,q^2$.
Since the quadratic equation does not change,sum of the roots,product of the roots for new equation is equal to the previous equation values.
Product of the roots, $pq=(pq{)}^2$
$\Rightarrow pq(pq-1)=0$
$\Rightarrow pq=1......\left(1\right)$
$\Rightarrow pq=0.......\left(2\right)$
Sum of the roots, $p+q=$$p^2+q^2$
$\Rightarrow (p+q{)}^2-2pq-(p+q)=0$
$\Rightarrow (p+q)\left(\right(p+q)-1)-2pq=0$
from (2) $pq=0$
$\Rightarrow p+q=0$....(3)
$\Rightarrow p+q=1$ .......(4)
Solving (2) & (3)
$p=0,q=0$
solving (2) & (4)
$pq=0,p+q=1$
$\Rightarrow p=0,q=1$ and $p=1,q=0$
from (1) $pq=1$
$\Rightarrow (p+q)\ast (p+q-1)=2$
$(p+q{)}^2-(p+q)-2=0$

$(p+q)=\frac{1\pm \sqrt{(}9)}{2}$

$(p+q)=\frac{1\pm 3}{2}$
$p+q=2----\left(5\right),p+q=-\mathrm{1....}\left(6\right)$
on solving $p+q=2,pq=1$
$p=1,q=1$
on solving $p+q=-1,pq=1$
$p^2+p+1=0$

$p=$$\frac{1\pm \sqrt{1-4}}{2}=\frac{1\pm i\sqrt{3}}{2}=\omega ,(\omega {)}^2$
$p=\omega ,q=(\omega {)}^2$ and $p=(\omega {)}^2,q=\omega$
$Hence(p,q)\to (0,0),(0,1),(1,0),(1,1),$ $(\omega ,(\omega {)}^2),((\omega {)}^2,\omega )$
$\therefore$ Probability to have equal roots$=$$\frac{2}{6}$$=$$\frac{1}{3}$