TS EAMCET · Maths · Quadratic Equation
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 \(143\)
Answer & Solution
Correct Answer
(D) \(143\)
Step-by-step Solution
Detailed explanation
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…
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