Challenge problem:

Given that http://www.mathhelpforum.com/math-he...f3f41408-1.gif with the equation having two unequal real roots,

Prove:http://www.mathhelpforum.com/math-he...0dc8787b-1.gif

moderator approved CB

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- June 5th 2010, 09:15 AMalexandrosquadratic equation
Challenge problem:

Given that http://www.mathhelpforum.com/math-he...f3f41408-1.gif with the equation having two unequal real roots,

Prove:http://www.mathhelpforum.com/math-he...0dc8787b-1.gif

moderator approved CB - June 5th 2010, 11:09 AMDeadstar
I've probably done something wrong but here goes...

Using standard formula we get roots to be

.

Since roots are real and unequal,

Hence...

.

Canceling we get

...

Hence result. - June 5th 2010, 03:13 PMmr fantastic
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- June 5th 2010, 03:47 PMCaptainBlack
- June 5th 2010, 04:05 PMmr fantastic
- June 5th 2010, 06:16 PMDeadstar
- June 6th 2010, 04:50 PMalexandros
If i understand you correctly :from you get ,by raising 1st the inequality to the power of 2 and then multplying by 16.

But when the inequality is raised to 2 must we not have that b>0 ,otherwise the inequality may change if b<0.The problem assumes only .

Example:take b=-3 and a=1 :we have ,but - June 6th 2010, 06:05 PMDeadstar
Yeah you're right.

Split it up into two cases then.

we know.

So let case 1 be

which was my first post and hence .

Case 2 is

and hence .

From which we get that...

Although I haven't checked I think some of the the > can be replaced with for one of the cases to include with . (This will only happen when b is negative since if b is +ve then ) - June 7th 2010, 03:28 AMalexandros