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

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

Grade:11

5 Answers

Sargun Nagpal
20 Points
9 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
9 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
9 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
askIITians Faculty 74 Points
9 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\leq0
i.e. b2-20a\leq0
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
15 Points
4 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  \small \frac{b^{2}}{4} + b = f
\small \frac{\mathrm{d} f}{\mathrm{d} x} = 0
\small \frac{b}{2} + 1 = 0
b= -2
 
So Minimum Value of 5a + b is,
Min value of   \small \frac{b^{2}}{4} + b, at b=-2
 
\small \mathbf{ANS}\; \mathbf{-1}
Minimum Value of 5a + b = -1

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