Question

‘p’ and ‘q’ are the zeroes of the polynomial
$a{x}^2+bx+c$. The values of $a$ and $b$ are

• A. -1, -(p + q)
• B. 1, p + q
• C. 1, -(p+q)
• D. -1, -(p+q)

Answer 1

The correct option is C 1, -(p+q)

Since, ‘p’ and ‘q’ are the zeroes of the polynomial $a{x}^2+bx+c$,

(x-p) and (x-q) are the factors of the expression $a{x}^2+bx+c$,

so $a{x}^2+bx+c=(x-p)(x-q)$

$⟹a{x}^2+bx+c=x^2-(p+q)x+pq$

By comparing the co-efficients of $x^2$, $x$, and the constant term, we get

$a=1$, $b=-(p+q)$ and $c=pq$