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irspow
2005-12-02 08:18:21

If you look inside the brackets of the numerator the sum within them is raised to the one-half power which is the same thing as taking the square root but just different notation.

RickyOswaldIOW
2005-12-01 12:50:29

I've been using the quadratic formula to solve equations on which I first complete the square.  I notice you are using the same formula here but without the sqrt?

RickyOswaldIOW
2005-11-30 09:09:06

Wow thanks, I'll make my answer a little more specific next time.

Again, Thanks

irspow
2005-11-30 08:34:24

I would personally use the quadratic equation for this problem:

-x^2 + 4x + 5

The quadratic equation would yield the same thing without trying different combinations.

[ -b +/- (b^2 - 4ac)^(1/2)] / 2a   =  [-4 +/- (16 + 20)^(1/2)] / -2, so x = 5 and -1

Maybe I am biased, but this method always seemed easier than figuring out values by trial and error.

Your question though was specifically about the "textbook" method of factoring.  I would describe it as follows.

You want your answer to be like: (ax + b)(cx + d).  This equals (ac)x^2 + (ad + bc)x + (bd).
So you are really concerned with the coefficients in the original equation.

In this case: ac = -1,  ad + bx = 4, and bd = 5.  b and d are the second terms, so just ask yourself what two numbers multiply to produce 5.  The simple answer leaves only one and five.    (and if you later find that you need fractional coefficients, God help you.)

(    + 1)(     + 5).......they both must be postitive otherwise the constant would be negative.

As ac = -1 and a and c represent the first terms just plug in a one into both brackets. Remember that one of these terms must be negative!

(1x  +1)(1x  +5)

The signs within the brackets come from the ad + bc term that produces the x^1 term in your original equation.  So ad = 5 and bc = 1 and these two answers must add up to 4.  The only way that a 5 and a 1 can equal 4 is if the 1 is negative.  Therefore the bc term must be negative and b is already a positive one so c must be negative.

(x+1)(-x+5)

Remember the (ac)x^2 + (ad + bc)x + bd form of the equation to keep all of this straight in the harder factoralizations.

Now you see why I almost always use the quadratic equation to find the zeros of a 2nd degree equation.  Just plug in the coefficients and your done.

RickyOswaldIOW
2005-11-30 06:06:22

I know the answer was right, I looked it up
My question (which I realise is not that clear) was,  "How do I work it out?"

mathsyperson
2005-11-30 06:01:40

Factorising -x² + 4x + 5 gives you (x+1)(5-x), so the roots would be -1 and 5.