Multiple choice

If $ a+b+c > 0 $, $a < 0 < b < c$, then roots of the equation $\displaystyle a\left ( x-b \right )\left ( x-c \right )+b \left ( x-c \right )\left ( x-a \right )+c\left ( x-a \right )\left ( x-b \right )=0 $ are real.

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Both Assertion and Reason are incorrect
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A Correct answer
AI explanation

Expanding the given equation a(x-b)(x-c) + b(x-c)(x-a) + c(x-a)(x-b) = 0 gives (a+b+c)x^2 - (ab+bc+ca)x + 3abc = 0. Because a+b+c is greater than 0, substituting x = a, x = b, and x = c into the quadratic expression yields values of 3a(b-a)(c-a), -3b(a-b)(c-b), and 3c(a-c)(b-c) respectively. Given the condition a < 0 < b < c, f(a) is negative, f(b) is positive, and f(c) is positive, which means the curve crosses the x-axis between a and b. As x tends to infinity, the function tends to positive infinity, guaranteeing a second real root, so the assertion is correctly explained by this reasoning.

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