Factoring a quadratic: strange issues

[ducmod]

Homework Statement

Hello!

One of the easiest rules (when possible to apply) to factor a quadratic is to find both x-s by

x1 + x2 = b
x1 * x2 = c

Homework Equations

Please, take a look at what is written in the book. I can't grasp why x1 = -2 and x2 = 3, and not, as I thought, x1 = -3 and x2 = 2.

The Attempt at a Solution

What is wrong in my understanding?

x1 + x2 = -1
x1 * x2 = -6
Therefore, x1 = -3 and x2 = 2; and the equation (x + 3 )(x - 2) =0

I would be grateful for the help.
Thank you!

 

[epenguin]

Homework Statement

Hello!

One of the easiest rules (when possible to apply) to factor a quadratic is to find both x-s by

x1 + x2 = b
x1 * x2 = c

Homework Equations

Please, take a look at what is written in the book. I can't grasp why x1 = -2 and x2 = 3, and not, as I thought, x1 = -3 and x2 = 2.

The Attempt at a Solution

What is wrong in my understanding?

x1 + x2 = -1
x1 * x2 = -6
Therefore, x1 = -3 and x2 = 2; and the equation (x + 3 )(x - 2) =0

I would be grateful for the help.
Thank you!

Can see only two lines of your quote so not sure what your problem is, but could it be this
x1 + x2 = b

Which should be -b. ?

 

[ducmod]

Can see only two lines of your quote so not sure what your problem is, but could it be this
x1 + x2 = b

Which should be -b. ?

Thank you for your reply. There is a screenshot attached, and my equations.
Do you say that x1 + x2 equal -b and not b? Mathisfun lied to me )))

 

[epenguin]

Thank you for your reply. There is a screenshot attached, and my equations.
Do you say that x1 + x2 equal -b and not b? Mathisfun lied to me )))

Check that again, but better and faster would be to look see if roots are a and b so that they satisfy

(x - a)(x - b) = 0

what is the coefficient of x in the quadratic you get expanding that?

 

[SteamKing]

Thank you for your reply. There is a screenshot attached, and my equations.
Do you say that x1 + x2 equal -b and not b? Mathisfun lied to me )))

This page from Math is Fun gives the correct relationship for the sum and product of the roots of a quadratic:

http://www.mathsisfun.com/algebra/polynomials-sums-products-roots.html

You can always check with Wikipedia for quadratic equations.

BTW, these formulas are called Vieta's formulas, for the famous French mathematician.

 

[HallsofIvy]

You can multiply the left side to see that $(x+ a)(x+ b)= x^2+ (a+ b)x+ ab$. In your given example, $x^2- x- 6$, so we must have a+ b= -1. And ab= -6. Obviously, 3 and -2 multiply to give -6. But -2+ 3= 1, not -1. We must have a=- 2 and b= 3. Once we have that we have the two factors (x+ (-2)(x+ 3)= (x- 2)(x+3)= 0, we must have either x- 2= 0 or x+ 3= 0. From those, x= 2 and x= -3 are the roots of the equation $x^2- x- 6= 0$.

 

[Dr. Courtney]

There are several ways to find the roots of a quadratic:

Factoring
Quadratic formula
Completing the square
Graphing

If a problem requires factoring and you can't figure it out that way, you can always use another method to find the roots and work backwards.

Suppose you use the quadratic formula to find roots f and g. The factors are then:

(x - f) and (x - g).