Question

If the zeroes of the quadratic polynomial $x^2+(a+1)x+b$ are $2$ and $3$, then

• A. $a=7$, $b=1$
• B. $a=5$,$b=1$
• C. $a=2$, $b=6$
• D. $a=-6$,$b=6$

Answer 1

The correct option is D $a=-6$,$b=6$

For the given polynomial:

$x^2+(a+1)x+b$,

the roots of the equation are $2,3$

Thus,

$2^2+(a+1)2+b=0$
$2a+b+6=0$ ...(i)
And,

$3^2+(a+1)3+b=0$
$3a+b+12=0$......(ii)
Subtracting (i) from (ii)
$a+6=0$
$a=-6$

Put this value in either (i) or (ii), we get $b=6$