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# Linear Equation question

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Intern
Joined: 15 Oct 2017
Posts: 28
Schools: Northeastern '20

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19 Mar 2018, 02:55
Hello,

Steep 1 : x²-5x-6=0

Steep 2 : (x-6)(x+1)

How should we make the transition? I don't get it

Thanks!
PS Forum Moderator
Joined: 16 Sep 2016
Posts: 314
GMAT 1: 740 Q50 V40

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20 Mar 2018, 10:38
thegame12 wrote:
Hello,

Steep 1 : x²-5x-6=0

Steep 2 : (x-6)(x+1)

How should we make the transition? I don't get it

Thanks!

Hi thegame12,

Step 2 is just factoring the given quadratic expression from step 1.

You can think like this...
(x - a)*(x - b) = x*x + x*(-b) + x*(-a) + a*b
= x^2 - (a+b)x +ab

You can remember the above as a result. Linear term is always sum of roots and constant term is the product.

So for given expression a+b=5 & a*b =-6
This is true for a = 6 and b = -1. ( You can tell this by trying out small numbers )

So step 2: (x - 6)*(x - (-1)) = (x-6)*(x+1)

Does that make sense?

Best,
Math Expert
Joined: 02 Sep 2009
Posts: 50060

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22 Mar 2018, 03:11
Manhattan Prep Instructor
Joined: 04 Dec 2015
Posts: 629
GMAT 1: 790 Q51 V49
GRE 1: Q170 V170

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05 Apr 2018, 10:15
thegame12 wrote:
Hello,

Steep 1 : x²-5x-6=0

Steep 2 : (x-6)(x+1)

How should we make the transition? I don't get it

Thanks!

You want to find two numbers - let's call them A and B - that have the following properties:

- when you multiply them together, you get the last number in the quadratic (in this case, -6. Be careful about the negative sign!)
- when you add them together, you get the middle number in the quadratic (in this case, -5.)

So, what two numbers add together to make -5, and multiply together to make -6?

The easiest place to start is to think about which numbers multiply to -6, then try adding them together and see what happens.

For instance, -3*2 = -6, but -3+2 doesn't equal -5. So that isn't the right pair of numbers.

But, -6*1 = -6, and -6+1 = -5. That's the right pair of numbers.

Once you have those numbers, you can write the quadratic like this:

(x-6)(x+1) = 0

Again, be careful to keep the negative signs the same as what you figured out earlier!

Finally, two things to be careful about:
- you can only use this technique when the first term in the quadratic doesn't have a coefficient. In other words, you can only do this when the quadratic starts with just $$x^2$$ (or any variable squared), not when it starts with something like $$2x^2$$. If it does, you have to divide the whole quadratic to make the 2 go away first.
- this technique is only useful when the quadratic is equal to 0. That is, when there's a 0 on the other side of the equals sign. The whole point of doing this is to find the values of x, and it's only easy to do that when you know that the product comes out to 0.
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Chelsey Cooley | Manhattan Prep Instructor | Seattle and Online

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Intern
Joined: 16 Oct 2011
Posts: 13

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10 Apr 2018, 08:14
if a*b = 0 then (1) a=0 (2) b=0, or both a=0 and b=0. We can set each binomial factor as a linear equation. x-6=0-->x=6, x+1=0--->x=-1
Re: Linear Equation question &nbs [#permalink] 10 Apr 2018, 08:14
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