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FMathsStudent

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Roots of Quadratic Equations - Proving

Hey, there are two questions that I am a little stuck on.

1/ One root of the equation ax²+bx+c=0 is the reciprocal of the other. Prove that c=a.

I know that the sum of the roots, α and 1/α:

α + 1/α = (α²+1)/α

And that the product of the roots would be:

α x 1/α = 1

From the roots bit, we can say that 1 = c/a. And therefore c=a.

Is that a correct proof? Or do I also need to use the sum of the roots in the equation?

2/ The roots of the equation ax²+bx+c=0 differ by 1. Prove that b²-a²-4ac=0

I know that the sum of the roots, α and α+1:

α+α+1 = 2α+1

2α+1=-b/a
2α = (-b-a)/a
α = (-b-α)/2a

And that the sum of the roots would be:

α x α+1 = α²+α

α²+α=c/a

[(-b-α)/2a]²+(-b-α)/2a=c/a

And then I get stuck after that...

Because after all the multiplying out and stuff, I get

2a²+2b²+4ab-4ac=0

Please help me and thank you in advance. ^_^

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Roots of Quadratic Equations - Proving

I'd do questions like this by writing the quadratic equation in terms of its roots, multiplying out and equating coefficients with the abc form.

So for 1), you know that the two roots are α and 1/α.
Therefore, the equation looks like k(x-α)(x-1/α).
This is kx² - k(α+1/α)x +kα(1/α) = kx² - k(α+1/α) +k

The quadratic term and the constant term are both k, and so a=c.

The second question can be solved using the same trick.

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Why did the vector cross the road?
It wanted to be normal.

JaneFairfax

Member

Registered: 2007-02-23

Posts: 6,868

Re: Roots of Quadratic Equations - Proving

[(-b-α)/2a]²+(-b-α)/2a=c/a

And then I get stuck after that...

Because after all the multiplying out and stuff, I get

2a²+2b²+4ab-4ac=0

Please help me and thank you in advance. ^_^

is correct. If you multiply it out correctly, you should get

.

Check your working carefully.

mathsmypassion

Member

Registered: 2008-12-01

Posts: 33

Re: Roots of Quadratic Equations - Proving

First piece of advice: try not to use "a" for the root if you have an "a" as coefficient in your equation. It is confusing.
Second piece of advice: The problem can be solved using the sum and the product of the roots. The idea is that the relationship you use must be simmetrical in x1 and x2 (the roots).

So we know that x1+x2 = -b/a and x1x2=c/a
x1-x2 is not simmetrical in x1 and x2 but (and here is the trick) (x1-x2)²  is.
(x1-x2)² =x1² +x2² -2x1x2 = (x1+x2)² -4x1x2= (-b/a)² -4(c/a)= 1
or b² /a² -4c/a-1=0 or b² - 4ac-a² =0 QED
I hope it will help you understand more about the sum and product of the roots.