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## Comment on

Factoring - Quadratics## This is one of the questions

Which of the following equations has

a root in common with x^2 - 6x + 5 =

0?

(A) x^2 + 1 = 0

(B) x^2 - x - 2 =0

(C) x^2 - 10x - 5 =0

(D) 2x^2 - 2 =0

(E) x^2 - 2x - 3 =0

I understood the solution but I have a question:

How X^2+1=0? Is this even possible? X^2 could be either zero or a positive number.

## Question link: http://www

Question link: http://www.beatthegmat.com/common-root-t117001.html

Great question, Mohammad!

Not all equations have a solution. This equation (x^2 + 1 = 0) is one of them.

ASIDE: some students with an understanding of complex (aka imaginary) numbers, will say that there IS a solution. It's x = i (where i = √-1). However, the GMAT only covers REAL numbers, so we say that the equation x^2 + 1 = 0 has no REAL solution.

## Thanks. And great point on

## Hi Brent,

Great video! Just a quick question - how does one go about factoring a quadratic equations such as these ones:

1) 2x^2 + 9x + 9 = 0

This one factors out to be (2x + 3) (x + 3)

2) 2x^2 + x - 3

Factors out to (x - 1) (2x + 3)

I'm confused on how to separate factors such as 2x in individual expressions in the above examples.

Hope this question made sense!

## Hi BalysLTU,

Hi BalysLTU,

There's a formal technique for factoring quadratics where the coefficient of the x² is NOT 1, but for the purposes of the GMAT, we can typically apply the informal method describe below.

Here's a video of the formal technique: https://www.khanacademy.org/math/algebra/polynomial-factorization/factor...

On the GMAT, you'll typically see quadratic equations that look like the following:

x² - 2x - 15 = 0

x² + 8x + 12 = 0

x² - 7x + 10 = 0

However, if you do encounter a quadratic equation where the coefficient of the x² is NOT 1, then you can typically factor the expression by applying some number sense and testing some values.

Here's what I mean:

Given: 2x² + 9x + 9 = 0

Let's say the expression (2x² + 9x + 9) can be factored to look something like: (a + b)(c + d)

What can we conclude about some of the values?

Well, we know that ac = 2x² (applying the FOIL method)

If we limit ourselves to integer vales, then there's only one way to get a product of 2x²

That is, (2x)(x) = 2x²

So, we already know that 2x² + 9x + 9 = (2x + b)(x + d)

We also know that bd = 9

There aren't that many options where the product of two integers equals 9

One option is (1)(9) = 9

Another option is (3)(3) = 9

At this point, we can start TESTING some options...

How about b = 1 and d = 9

Plug in to get: (2x + 1)(x + 9)

Expand and simplify to get: (2x + 1)(x + 9) = 2x² + 19x + 9

The middle term (19x) is not right. We want 9x

TRY AGAIN

How about b = 3 and d = 3

Plug in to get: (2x + 3)(x + 3)

Expand and simplify to get: (2x + 3)(x + 3) = 2x² + 9x + 9

Perfect!

Done!

I'll leave it to you to try factoring the second expression.

Cheers,

Brent

## Hey!

How do you factorize x² - 10x - 5? I cant seem to think of two numbers that will add to be - 10 and multiply to be 5. Is it because it consists of two negatives? (Since its not x² + nx + p). I'm confused what to do when both the x term and the constant are negative, as opposed to two positives or one positive and one negative. Cheers!

## Good question, Byefox.

Good question, Byefox.

x² - 10x - 5 cannot be factored into the form (x + ?)(x - ?)

Can you tell me the details of the question that require the factorization of x² - 10x - 5? Perhaps there's an alternate approach.

Cheers,

Brent

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