Question:
If $p$ and $q$ are the roots of the equation $x^{2}-p x+q=0$, then
(a) $p=1, q=-2$
(b) $b=0, q=1$
(c) $p=-2, q=0$
(d) $p=-2, q=1$
Solution:
Given that $\rho$ and $q$ be the roots of the equation $x^{2}-p x+q=0$
Then find the value of $p$ and $q$.
Here, $a=1, b=-p$ and, $c=q$
p and q be the roots of the given equation
Therefore, sum of the roots
$p+q=\frac{-b}{a}$
$=\frac{-p}{1}$
$=-p$
$q=-p-p$......(1)
$=-2 p$
Product of the roots
$p \times q=\frac{q}{1}$
As we know that
$p=\frac{q}{q}$
$=1$
Putting the value of $p=1$ in equation (1)
$q=-2 \times 1$
$=-2$
Therefore, the value of $p=1 ; q=-2$
Thus, the correct answer is (a)