The set of all values of k for which the quadratic equation x squared plus 2(k minus 1)x plus k plus 5 equals 0 has roots of opposite signs is

A. k < -5 — For a quadratic equation to have roots of opposite signs, the product of the roots must be negative. Since the product of the roots is c divided by a, we set k plus 5 less than 0, which yields k less than -5.

JEE Main · MathematicsMedium

$The set of all values of k for which the quadratic equation x^{2} + 2 (k - 1) x + k + 5 = 0 has roots of opposite signs is$

A. $k < - 5$
B. $k > 1$
C. $- 5 < k < 1$
D. $k < - 5 or k > 1$

Show correct answer & step-by-step solution

Correct answer: A — $k < - 5$

Solution

For roots of opposite signs, the product of roots P = \frac{c}{a} < 0 . Here P = k + 5, so k + 5 < 0 implies k < - 5 .

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The set of all values of k for which the quadratic equation x2+2(k−1)x+k+5=0 has roots of opposite signs is

Solution

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More Quadratic Equations practice questions

$The set of all values of k for which the quadratic equation x^{2} + 2 (k - 1) x + k + 5 = 0 has roots of opposite signs is$