Question

The value of the expression x2 +2bx+cis positive,
if -
(A) b^2-4c<0 (B)b^2-4c>0
(C) c^2 <b
(D) b^2 <C​

Answer 1

Step-by-step explanation:

If the discriminate is greater then 0 then the equation is positive and have real and distinct roots .....

Answer 2

Answer: The value of the expression x2 +2bx+c is positive if b² < c

Step-by-step explanation:

Let the expression x² + 2bx + c be p(x)

p(x) = x² + 2bx + c

p(x) represents an upward opening parabola as the coefficient of x² is 1 which is greater than 0.

p(x) will have its minimum value at its vertex.

If we constrain the minimum value to be greater than zero, then p(x) will be positive for all values of x.

At vertex, slope of tangent is 0.

⇒   $\frac{dp(x)}{dx} = 0$

⇒  $\frac{d(x^{2} + 2bx+c)}{dx} = 0$

⇒ $2x + 2b = 0$

⇒ $x = -b$

Value of p(x) at x = -b  ⇒ p(-b)

p(x) = x² + 2bx + c

p(-b) = (-b)² + 2b(-b) + c

p(-b) =  c - b²

p(-b) should be greater than zero

∴ c - b² > 0

⇒ c > b²

⇒ b² < c

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