# Math Help - Finding equation from roots

1. ## Finding equation from roots

I am trying to prepare for this math course and get to realize that most of my problem in doing math is understanding the question. there are a set of problem that involves writing and simiplifying an equation from the roots of a given equation. I tried solving and ... just not getting the correct anwer.
Here is one 3x^2+2x-1=0, and the new equation's roots are the reciprocal of the one I just gave and must be simplified.
There is another with an equation 4x^2-5x-2 and the same condition only this time the new equation is double those of the given. Can some one help me with problem?

2. Originally Posted by scrible
I am trying to prepare for this math course and get to realize that most of my problem in doing math is understanding the question. there are a set of problem that involves writing and simiplifying an equation from the roots of a given equation. I tried solving and ... just not getting the correct anwer.
Here is one 3x^2+2x-1=0, and the new equation's roots are the reciprocal of the one I just gave and must be simplified.
There is another with an equation 4x^2-5x-2 and the same condition only this time the new equation is double those of the given. Can some one help me with problem?
Do you know how to find the roots of the equation you were given? With the quadratic formula:

If:

$ax^2+bx+c=0$

Then:

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

???

3. Originally Posted by scrible
I am trying to prepare for this math course and get to realize that most of my problem in doing math is understanding the question. there are a set of problem that involves writing and simiplifying an equation from the roots of a given equation. I tried solving and ... just not getting the correct anwer.
Here is one 3x^2+2x-1=0, and the new equation's roots are the reciprocal of the one I just gave and must be simplified.
There is another with an equation 4x^2-5x-2 and the same condition only this time the new equation is double those of the given. Can some one help me with problem?
$3x^2+2x-1 = 0$

$(3x-1)(x+1) = 0$

$x = \frac{1}{3}$ , $x = -1$

$4x^2-5x-2 = 0$

$x = \frac{5 \pm \sqrt{57}}{8}$

4. Originally Posted by skeeter
$3x^2+2x-1 = 0$

$(3x-1)(x+1) = 0$

$x = \frac{1}{3}$ , $x = -1$

$4x^2-5x-2 = 0$

$x = \frac{5 \pm \sqrt{57}}{8}$

The answer that I got was $x^2+2x-3$ but the book had $x^2-2x-3$. How did they get that?

5. Originally Posted by scrible
The answer that I got was $x^2+2x-3$ but the book had $x^2-2x-3$. How did they get that?
roots of $y = 3x^2 + 2x - 1$ are $x=\frac{1}{3}$ and $x=-1$

reciprocal roots are $x=3$ and $x=-1$

$(x-3)(x+1) = x^2 - 2x - 3$

6. Skeeters method, find the roots to the original equation, take the reciprocal, and create the new equation from that is the obvious and simplest method.

Here is a little more "sophisticated" method. Look at the general product (x- a)(x- b)= x^2- (a+b)x+ ab. If, instead, we have (x- 1/a)(x- 1/b) we get x^2- (1/a+ 1/b)x+ 1/ab. You can see that the last term, 1/ab, is just the reciprocal in the original equation, ab. Also 1/a+ 1/b= b/ab+ a/ab= (a+b)/ab so the coefficient of x in the last equation is the coeffient of x in the original equation divided by the last term.

Starting from 3x^2+2x-1=0, which is the same as x^2+ (2/3)x- 1/3= 0 (I divided both sides by 3), we can see that the equation having the reciprocals as roots is x^2+ (2/3)/(-1/3)x- 1(1/3)= x^2- 2x- 3= 0.

We can do the same kind of thing for "2a" and "2b". (x- 2a)(x- 2b)= x^2-(2a+2b)+ (2a)(2b)= x^2- 2(a+b)+ 4(ab)= 0. To create an equation that has 2 times the roots as its roots, multiply the coefficient of x by 2 and the final term by 4.

Whatever the roots of 4x^2-5x-2= 0, which is the same as x^2- (5/4)x- 1/2= 0, are, the roots of the equation x^2- (5/2)(2)x- (1/2)(4)= x^2- 5x- 2= 0 are twice that.

7. Originally Posted by HallsofIvy
Skeeters method, find the roots to the original equation, take the reciprocal, and create the new equation from that is the obvious and simplest method.

Here is a little more "sophisticated" method. Look at the general product (x- a)(x- b)= x^2- (a+b)x+ ab. If, instead, we have (x- 1/a)(x- 1/b) we get x^2- (1/a+ 1/b)x+ 1/ab. You can see that the last term, 1/ab, is just the reciprocal in the original equation, ab. Also 1/a+ 1/b= b/ab+ a/ab= (a+b)/ab so the coefficient of x in the last equation is the coeffient of x in the original equation divided by the last term.

Starting from 3x^2+2x-1=0, which is the same as x^2+ (2/3)x- 1/3= 0 (I divided both sides by 3), we can see that the equation having the reciprocals as roots is x^2+ (2/3)/(-1/3)x- 1(1/3)= x^2- 2x- 3= 0.

We can do the same kind of thing for "2a" and "2b". (x- 2a)(x- 2b)= x^2-(2a+2b)+ (2a)(2b)= x^2- 2(a+b)+ 4(ab)= 0. To create an equation that has 2 times the roots as its roots, multiply the coefficient of x by 2 and the final term by 4.

Whatever the roots of 4x^2-5x-2= 0, which is the same as x^2- (5/4)x- 1/2= 0, are, the roots of the equation x^2- (5/2)(2)x- (1/2)(4)= x^2- 5x- 2= 0 are twice that.

There is another one that I am getting trouble with and it said to write down the equation whose roots are minus of $2x^2-3x-1$ I tried $(x-(a-1))(x-(B-1))$?