Algebra> If ax2 - bx +5 = 0 does not have 2 distin...

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

Gyan Prakash Karn , 10 Years ago

Grade 11

5 Answers

Sargun Nagpal

Last Activity: 10 Years ago

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

Last Activity: 10 Years ago

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

Last Activity: 10 Years ago

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

Last Activity: 10 Years ago

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