1. ## Question on roots of a quadratic polynomial.

Check out the image . Question number 5

Pls give me the solution...

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2. ## Re: Question on roots of a quadratic polynomial.

Do you know the "quadratic formula"? What does the quadratic formula give for the two solutions to $\displaystyle x^2+ ax+ b= 0$? What is the difference between them? What does the quadratic formula give for the two solutions to $\displaystyle x^2+ bx+ a= 0$? What is the difference between them? Set those two differences equal. What does that give you?

3. ## Re: Question on roots of a quadratic polynomial.

Originally Posted by HallsofIvy
Do you know the "quadratic formula"? What does the quadratic formula give for the two solutions to $\displaystyle x^2+ ax+ b= 0$? What is the difference between them? What does the quadratic formula give for the two solutions to $\displaystyle x^2+ bx+ a= 0$? What is the difference between them? Set those two differences equal. What does that give you?
Sryi didnt get it

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4. ## Re: Question on roots of a quadratic polynomial.

Originally Posted by sbjsavio
Check out the image . Question number 5

Pls give me the solution...
Do you expect us to stand on our heads?
Do you really want help?

5. ## Re: Question on roots of a quadratic polynomial.

Originally Posted by Plato
Do you expect us to stand on our heads?
Do you really want help?

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6. ## Re: Question on roots of a quadratic polynomial.

Originally Posted by sbjsavio
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Well good luck $\sigma\mu\alpha\rho\tau~\alpha\sigma\sigma$.

7. ## Re: Question on roots of a quadratic polynomial.

Ok sorry for those silly things.
Thank u guys i got it by solving.I used that quadratic formula method from hallofivy.
Gud luck..

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8. ## Re: Question on roots of a quadratic polynomial.

The roots of $\displaystyle x^2+ax+b=0$ are:

$\displaystyle x=\frac{-a\pm\sqrt{a^2-4b}}{2}$

Taking the difference $\displaystyle \Delta r_1$ to be the larger root minus the smaller, we have:

$\displaystyle \Delta r_1=\sqrt{a^2-4b}$

Likewise, for the second quadratic we would find:

$\displaystyle \Delta r_2=\sqrt{b^2-4a}$

If these differences are the same, this implies:

$\displaystyle a^2-4b=b^2-4a$

Can you proceed?

9. ## Re: Question on roots of a quadratic polynomial.

Originally Posted by MarkFL
The roots of $\displaystyle x^2+ax+b=0$ are:

$\displaystyle x=\frac{-a\pm\sqrt{a^2-4b}}{2}$

Taking the difference $\displaystyle \Delta r_1$ to be the larger root minus the smaller, we have:

$\displaystyle \Delta r_1=\sqrt{a^2-4b}$

Likewise, for the second quadratic we would find:

$\displaystyle \Delta r_2=\sqrt{b^2-4a}$

If these differences are the same, this implies:

$\displaystyle a^2-4b=b^2-4a$

Can you proceed?
I got the answer as a + b equals -4 and that is the correct answer given in my text books answer key. @MarkFL thank u brother

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