What is the discriminant for this quadratic?

[Misr]

Homework Statement

Prove that for all real values of a and b , the roots of the eqn : ax^2-(2a+b)x+b-5a=0
are real and different roots

Homework Equations

discriminat=b^2-4ac
where a is the x^2 coefficient and b is the x coefficient and c is the absolute term

The Attempt at a Solution

(2a+b)^2 - (4a(b-5a)) = 4a^2+b^2+4ab - 4ab+20a^2 = 4a^2+b^2+20a^2

so in order to solve the problem 4a^2+b^2+20a^2 should be > zero
and of course 4a^2+b^2+20a^2 >=0

but if a and b are zero then whole expression is going to be zero thus the roots are real but the same

so can u help me??
Thanks

 

[ehild]

If both a and b are zero, the original equation says that 0x^2+0x=0, an identity, which is fulfilled by any x. Mention this when you solve the problem and then say, that in any other cases the discriminant is positive (why?).

ehild

 

[Misr]

I don't really understand this.

that in any other cases the discriminant is positive (why?).

Because we use the square of a and b so can't be negative
my problem is when they are zero
Thanks

 

[ehild]

Sorry, silly me. I wanted to say that the original equation becomes 0*x^2 + 0*x =0, an identity, of which all numbers are roots. So there is not only one root in this case or two equal ones, but infinite. You are right, this case should have been excluded from the problem.

ehild

 

[Misr]

Do u mean that there's somethin wrong with the problem??
I think it should be : ax^2-(2a+b)x+b-5=0 instead of ax^2-(2a+b)x+b-5a=0
Right?

 

[ehild]

No, that equation would not have any roots for a=0 b=0. The problem should say, that "Prove that for all real, nonzero values of a and b a, the roots of the eqn : ax^2-(2a+b)x+b-5a=0 are real and different".

ehild

 

[Misr]

No, that equation would not have any roots for a=0 b=0

It would have infinite roots as you mentioned before right?

The problem should say, that "Prove that for all real, nonzero values of a and b a, the roots of the eqn : ax^2-(2a+b)x+b-5a=0 are real and different".

anyways there's sometythin wrong with the problem.
Thanks

 

[ehild]

I think it should be : ax^2-(2a+b)x+b-5=0 instead of ax^2-(2a+b)x+b-5a=0
Right?

I meant that ax^2-(2a+b)x+b-5=0 would not have any roots for a=0, b=0, as it would look: 0*x^2-0*x-5 =0, that is -5=0 which is false, there is no x that makes it true. The original equation has any number as root.

The formulation of the problem is wrong.

ehild