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## Comment on

Basic Equation Solving## Hey Brent can you help me

## If the quadratic is NOT in

If the quadratic is NOT in the form x² + bx + c = 0, then it is very likely going to be in the form of one of the 2 SPECIAL PRODUCTS:

(x + y)² = x² + 2xy + y²

(x - y)² = x² - 2xy + y²

Here are some examples of Special Products:

4x² + 4x + 1 = (2x + 1)²

25x² - 10x + 1 = (5x - 1)²

9x² + 42x + 49 = (3x + 7)²

Notice the pattern?

In the 1st example, 4x² = (2x)(2x), 1 = (1)(1), and 4x = (2)(2x)(1)

In the 2nd example, 25x² = (5x)(5x), 1 = (1)(1), and 10x = (2)(5x)(7)

In the 3rd example, 9x² = (3x)(3x), 49 = (7)(7), and 42x = (2)(3x)(7)

Notice that 9x² + 6x + 1 = 0 is also a special product.

9x² = (3x)(3x), 1 = (1)(1), and 6x = (2)(3x)(1)

So, we can factor to get: (3x + 1)² = 0

This means 3x + 1 = 0

So, 3x = -1

x = -1/3

For more on Special Products, watch https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi... (start video at 4:48)

Cheers,

Brent

## Hi Brent ,

for https://gre.myprepclub.com/forum/the-integers-x-and-y-are-greater-than-1-if-4x-7y-2148.html

I had approached in this way

746=2*2*3*7*3*3

and from there 2*2=4 and multipled with 3

i.e considereed 3 as x

7*9 - considered y as 9

I got the answer but is this a correct approach to follow

Thanks

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/the-integers-x-and-y-are-greater-than-1...

Yes, that approach works perfectly.

Cheers,

Brent

## can you explain to me the one

## Can you please provide a link

Can you please provide a link to the question. Otherwise it'll take me a long time to find the question you're referring to.

## It is the second question

## Question link: https:/

Question link: https://gre.myprepclub.com/forum/n-2k-3m-15913.html

Please tell me the values you used that caused the two quantities to be not equal, and I'll take a look.

## I apologize, I wrote the 3 as

## No problem. It happens to me

No problem. It happens to me all the time :-)

## Hey brent

In this Question you mentioned a useful property https://gre.myprepclub.com/forum/if-sqrt-4-5-5-square-14654.html

However what i did was cancel the cubes root and sq root to get 1 as answer can we not cancel sq root from numenator and denominator or do we have to solve the fraction below sq root and than cube it to reomve cube root?

## Question link: https://gre

Question link: https://gre.myprepclub.com/forum/if-sqrt-4-5-5-square-14654.html

Can you explain what you mean when you say "cancel the cubes root and sq root to get 1"?

Are you saying that, if cuberoot(a) x squareroot(b) = cuberoot(c) x squareroot(d)...

... then it must also be true that ab = cd

If that's what you were saying, then it's not true

Consider this example:

cuberoot(8) x squareroot(16) = cuberoot(64) x squareroot(4) [this equation is true]

However, when we apply your property, we get: 8 x 16 = 64 x 4, which is not true.

It's important to recognize that cuberoot(k) = k^(1/3)

And squareroot(k) = k^(1/2)

In order to cancel terms in an equation, you must be able to do so by performing the same operation on both sides of the equation.

So my question for you is: What operation are you performing on both sides of the equation to cancel the roots?