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

Sargun Nagpal
20 Points
7 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
20 Points
7 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
20 Points
7 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
7 years ago
Hello student,
Given equation is ax^2-bx+5=0.
Since the equation does not have 2 distinct real roots, therefore discriminant0
i.e. b2-20a0
20a b2
5a b2/4
5a+b(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
15 Points
2 years ago
Since the Quadratic Equation has no two Real roots,
So D
b2 - 20a
b2

5a >= $\frac{b^{2}}{4}$
Minimum value of 5a =  $\frac{b^{2}}{4}$

So Minimum value of 5a + b =
Minimum of  $\inline \small \frac{b^{2}}{4} + b$ = f
$\inline \small \frac{\mathrm{d} f}{\mathrm{d} x}$ = 0
$\inline \small \frac{b}{2} + 1 = 0$
b= -2

So Minimum Value of 5a + b is,
Min value of   $\inline \small \frac{b^{2}}{4} + b$, at b=-2

$\inline \small \mathbf{ANS}\; \mathbf{-1}$
Minimum Value of 5a + b = -1