Finding the roots of px^2 + ax + 1 = 0 in terms of

M

Math wiz ya rite 09

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Aug 27, 2006
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Hello all!

The roots of an equation:

px^2 + qx + 1 = 0 are r sub1 and r sub2.

Find (r sub1)^2 + (r sub2)^2 in terms of p and q.

Thank you in advance,

M.W. 09
 
I believe you mean pX^2+qX+1?
Use the quadratic formula to get X=(-q+sqrt(q^2-4p))/(2p)
There are two values of X (notice the +sign) so call them r1, and r2, and square X.
 
Yes it should be qx (i fixed it in my original post now)

I see the quadratice,but I still don't see the manipulation to (r1)^2 + (r2)^2
 
r1, and r2 are the roots, but I used X with two values, so replace the two values with r1, and r2 and rectangle. (or square.)
 
Ok I see what r1 and r2 are now, but my problem is with simplifying


[(-q+sqrt(q^2-4p))/(2p)]^2 + [(-q-sqrt(q^2-4p))/(2p)]^2
 
By writing it on paper, you'll see it's not nearly as copmlicated as it looks.
sqrt means square root sign.
 
Math wiz ya rite 09 said:
Hello all!

The roots of an equation:

px^2 + qx + 1 = 0 are r sub1 and r sub2.

Find (r sub1)^2 + (r sub2)^2 in terms of p and q.

Thank you in advance,

M.W. 09
Click to expand...

Here's a thought.....
If r<SUB>1</SUB> and r<SUB>2</SUB> are the roots of a quadratic equation in the form

ax<SUP>2</SUP> + bx + c = 0,

then r<SUB>1</SUB> * r<SUB>2</SUB> = c / a
and r<SUB>1</SUB> + r<SUB>2</SUB> = -b / a

So, for your equation,
r<SUB>1</SUB>*r<SUB>2</SUB> = 1 / p
and
r<SUB>1</SUB> + r<SUB>2</SUB> = - q/p

(r<SUB>1</SUB> + r<SUB>2</SUB>)<SUP>2</SUP> = (-q / p)<SUP>2</SUP>


and,
(r<SUB>1</SUB> + r<SUB>2</SUB>)<SUP>2</SUP> = (r<SUB>1</SUB>)<SUP>2</SUP> + 2 r<SUB>1</SUB> r<SUB>2</SUB> + (r<SUB>2</SUB>)<SUP>2</SUP>

(-q / p)<SUP>2</SUP> = (r<SUB>1</SUB>)<SUP>2</SUP> + 2( 1/p) + (r<SUB>2</SUB>)<SUP>2</SUP>

Hopefully you can take that, and come up with (r<SUB>1</SUB>)<SUP>2</SUP> + (r<SUB>2</SUB>)<SUP>2</SUP>
 
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