Quadratic Eqn?

Question

If a equation is 0=[(-1x10^-5)(X^2)+0.6X]-H
The value of the equation will be min when H is maximum. So how to find the minimum possible value of H?

Rewriting the eqn:
0=-(10^-5)(X^2)+0.6X-H

I have been very careless!!!! The discriminant is zero when the value of H is maximum so simply 0=(0.6^2)+(4*-10^-5*H) which gives H=9000 which in fact is the correct ans!!!!!!

Answer 1

I am not really sure what you are asking but I would put the quadratic into vertex form to find the maximum of H.

I would first bring the h to the other side of the question and then complete the square on the remaining two terms.

H = -1x10^-5(x^2 - 60 000x)
H = -1x 10^-5[(x^2 - 60 000x + 30 000^2) -30 000^2]
H = -1 x 10^-5(x - 30 000)^2 + 9 000

Therefore the maximum value of H is 9 000 when x = 30 000.

Answer 2

then,
H = (-1x10^-5)X^2 + 0.6X

Take the derivative
dH/dx = 2(-1x10^-5)X + 0.6

Set it equal to zero, and calc X
(-2x10^-5)X + 0.6 = 0
X = -0.6 / -2x10^-5
X = 3 x 10^4

Max occurs at X = 3 x 10^4

You can test this by substituting a value for X that is slightly larger or slightly smaller than 30000.

Answer 3

Let
H=[(-1x10^-5)(X^2)+0.6X]
take derivative of H

H'=-2.10^-5 x + 0,6

Find x if H'=0

x=4. 10^4
This value makes H max.

So put this value in your function
(it must not be equation, if it were its value is ready as zero)

Result=-1. 10^-5 . (4. 10^4) +0,6. 4.10^4-4.10^4

result=4000,4

Answer 4

i myself have no ans
but with my limited knowledge

if we are able to make the equation a perfect square the remaing part would be the min valueof the whole equation when it has real roots

therefore
the value obtained is the max value of H