Question

The roots of the equation $x^2$ + x - (p +1) = 0, where p is a constant are

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

Answer 1

The correct option is

C

p,-(p+1)

The roots of the equation is $x^2$ + x - p (p + 1) = 0, where p is a constant.
Its solution can be solved by using quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
This can be done as
On comparing the given equation with $a{x}^2$+bx+c=0
a = 1, b = 1, c = -p(p+1)
$\therefore x=\frac{-1\pm \sqrt{1^2-4\times 1\times \{-p(p+1\left)\right\}}}{2\times 1}$
$x=\frac{-1\pm \sqrt{1-4(-p^2-p)}}{2}$
$=\frac{-1\pm \sqrt{(2p+1{)}^2}}{2}$
$=\frac{-1\pm (2p+1)}{2}$
$=\frac{-1+(2p+1)}{2}or\frac{-1-(2p+1)}{2}$
$=\frac{-1+(2p+1)}{2}=\frac{2p}{2}=p$
$=\frac{-1-(2p+1)}{2}=\frac{-2-2p}{2}=-1-p=-(p+1)$
Therefore, the roots are given by $x=p,-(p+1)$
The correct answer is C.