Question

The curves $y=x^{2}+9 x+20$ and $y=x^{2}+b x+c$ intersect the $X$-axis at the points $\left(\alpha_{i}, 0\right),(i=1,2,3,4)$. If $\alpha_{1} < \alpha_{2} < \alpha_{3} < \alpha_{4}$ be such that $\left|\alpha_{1}-\alpha_{3}\right|=\left|\alpha_{2}-\alpha_{4}\right|=8$, then the sum of all possible values of $b$ and $c$ is

• A. 186
• B. 159
• C. 216
• D. None of the above

Answer 1

Answer: (D)

The roots of quadratic equation
$x^{2}+9 x+20=0$ are $-4$ and $-5$
So, the possible roots of quadratic equation
$x^{2}+b x+c=0$ are according to given information
$\left(x_{1}, x_{2}\right)=(-13,4),(3,4),(-13,-12)$
$\therefore $ Possible values of $b$ are $9,-7,25$
and possible values of $c$ are $-52,12,156$
therefore sum of all possible value of $b$ and $c$ is
$9-7+25-52+12+156=143$