Q51 / 160 · TS EAMCET 2020 (09 Sep Shift 2) · Mathematics

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
    $216$
  • Correct answer
    D
    $143$

Solution

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 $9-7+25-52+12+156=143$

Hence, the correct answer is option: 'D'.

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