# Thread: Finding equation using roots

1. ## Finding equation using roots

Hi, I'm having trouble with this question: Find the equation whose roots are the squares of the reciprocals of the roots of x^2+6x-2.

So far, I've done:

Let x1 and x2 represent roots of the original equation.

x1 + x2= -b/a
= -6
(x1)(x2) = c/a
= -2

Then I figured the roots of the required equation would be (1/x1^2) and (1/x2^2)

So the product of the required equation would be 1/4. However, I'm having difficulty finding the sum. I think it should be:

x1^2 + x2^2
___________

4

However, I don't know how to fix the numerator so that I can sub a value into it and get the correct sum. (Then obviously, I can just sub the sum and product into x^2-x[sum]+product). I thought of square-rooting the entire equation, but that doesn't work out. Any help finding the sum for the required equation would be greatly appreciated Thank you!

2. Originally Posted by starswept
Hi, I'm having trouble with this question: Find the equation whose roots are the squares of the reciprocals of the roots of x^2+6x-2.

So far, I've done:

Let x1 and x2 represent roots of the original equation.

x1 + x2= -b/a
= -6
(x1)(x2) = c/a
= -2

Then I figured the roots of the required equation would be (1/x1^2) and (1/x2^2)

So the product of the required equation would be 1/4. However, I'm having difficulty finding the sum. I think it should be:

x1^2 + x2^2
___________

4

However, I don't know how to fix the numerator so that I can sub a value into it and get the correct sum. (Then obviously, I can just sub the sum and product into x^2-x[sum]+product). I thought of square-rooting the entire equation, but that doesn't work out. Any help finding the sum for the required equation would be greatly appreciated Thank you!
i don't remember the rules for the sum and products of roots and all that stuff, but i think i have another method that can work, if you're interested

or if you have to use that method, i'll look up those rules.

this site seems to be sufficient

3. Say that $x^2 + 6x - 2 = 0$ has roots p & q.
Then $p + q = - 6,\quad pq = - 2$ which implies $p^2 + q^2 = 36 - 2pq = 40.$
Now the equation you want is $\left( {x - \frac{1}{{p^2 }}} \right)\left( {x - \frac{1}{{q^2 }}} \right) = x^2 - \left( {\frac{{p^2 + q^2 }}{{p^2 q^2 }}} \right)x + \frac{1}{{p^2 q^2 }} = 0$.

Can you finish?

4. $\displaystyle \frac{1}{x_1^2}+\frac{1}{x_2^2}=\frac{x_1^2+x_2^2} {x_1^2x_2^2}=\frac{(x_1+x_2)^2-2x_1x_2}{(x_1x_2)^2}$

5. Originally Posted by Plato
Say that $x^2 + 6x - 2 = 0$ has roots p & q.
Then $p + q = - 6,\quad pq = - 2$ which implies $p^2 + q^2 = 36 - 2pq = 40.$
Now the equation you want is $\left( {x - \frac{1}{{p^2 }}} \right)\left( {x - \frac{1}{{q^2 }}} \right) = x^2 - \left( {\frac{{p^2 + q^2 }}{{p^2 q^2 }}} \right)x + \frac{1}{{p^2 q^2 }} = 0$.

Can you finish?
i was thinking of doing something like that. nice!

6. Originally Posted by red_dog
$\displaystyle \frac{1}{x_1^2}+\frac{1}{x_2^2}=\frac{x_1^2+x_2^2} {x_1^2x_2^2}=\frac{(x_1+x_2)^2-2x_1x_2}{(x_1x_2)^2}$
Just one question about this; how did you get (x1+x2)^2 - 2x1x2? That gives the correct answer, but I worked it out to be (x1+x2)^2 + 2x1x2. I'm just not certain how to get the minus 2x1x2.

7. Originally Posted by starswept
Just one question about this; how did you get (x1+x2)^2 - 2x1x2? That gives the correct answer, but I worked it out to be (x1+x2)^2 + 2x1x2. I'm just not certain how to get the minus 2x1x2.

$(x_1 + x_2)^2 = x_1^2 + 2x_1 x_2 + x_2^2$

$\Rightarrow x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2$