If ax2 - bx +5 = 0 does not have 2 distinct real roots, then find the minimum value of 5a + b?

Question

Sargun Nagpal · Answer

ax^2-bx+5=0. Since the equation does not have 2 distinct real roots, therefore discriminant i.e. b^2-20a 20a >/= b^2 5a >/= b^2 /4 5a+b >/= b^2/4 + b. b^2/4 + b is a quadratic equation whose minimum value is -1/ (2* 1/4) = -2. (Minimum value of ax^2+bx+c=0 where a>0 is -b/2a. Hence required ans = -2

Sargun Nagpal · Answer

ax^2-bx+5=0. Since the equation does not have 2 distinct real roots, therefore discriminant i.e. b^2-20a 20a >/= b^2 5a >/= b^2 /4 5a+b >/= b^2/4 + b. b^2/4 + b is a quadratic equation whose minimum value is -1/ (2* 1/4) = -2. (Minimum value of ax^2+bx+c=0 where a>0 is -b/2a .) Hence required ans = -2

Sargun Nagpal · Answer

Apologies. I am not able to edit the answer. The discriminant is less than or equal to zero, So b^2 – 20a is less than or equal to 0.

SHAIK AASIF AHAMED · Answer

Hello student,
Please find the answer to your question below
Given equation is ax^2-bx+5=0.
Since the equation does not have 2 distinct real roots, therefore discriminant $\leq$ 0
i.e. b²-20a $\leq$ 0
20a $\geq$ b²
5a $\geq$ b²/4
5a+b $\geq$ (b²)/4 + b.
(b²/4) + b is a quadratic equation whose minimum value is given by -1/(2*(1/4)) = -2.
(Minimum value of ax^2+bx+c=0 where a>0 is -b/2a .)
Hencethe minimum value of 5a + b is -2

Varun Maheshwari · Answer

Since the Quadratic Equation has no two Real roots, So D b 2 - 20a b 2 5a >= Minimum value of 5a = So Minimum value of 5a + b = Minimum of = f = 0 b= -2 So Minimum Value of 5a + b is, Min value of , at b=-2 Minimum Value of 5a + b = -1