If the roots of equation x 2-4 x - c -0 are real then A- C - 4B- C -4C- C - 4D- c -4

The correct option is C c ≤ 4Given, x^2 + 4x + c = 0 Value of discriminant is given by, nbsp; nbsp;b^2 4ac = 4^2 – 4c = 16 4c The roots of quadrati ...

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Standard X

Mathematics

Nature of Roots

If the roots ...

Question

If the roots of equation $x^{2} + 4 x + c = 0$ are real then

c \geq 4

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c > 4

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c = 4

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c \leq 4

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Solution

The correct option is D $c \leq 4$
Given, $x^{2} + 4 x + c = 0$

Value of discriminant is given by, $b^{2} - 4 a c = 4^{2} - 4 c = 16 - 4 c$

The roots of quadratic equation are real then discriminant $\geq 0 16 - 4 c \geq 0 - 4 c \geq - 16 4 c \leq 16$
(The inequality changes when the equation is multiplied by $^{'} -^{v e}^{'}$

$∴ c \leq 4$

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If the roots of equation x2+4x+c=0 are real then

The correct option is D c≤4Given, x2+4x+c=0 Value of discriminant is given by, b2−4ac=42–4c=16−4c The roots of quadratic equation are real then discriminant ≥016–4c≥0−4c≥−164c≤16 (The inequality changes when the equation is multiplied by ′−ve ′ ∴c≤4

If the roots of equation $x^{2} + 4 x + c = 0$ are real then