# Lesson: Basic Equation Solving

## Comment on Basic Equation Solving

### Hey Brent can you help me

Hey Brent can you help me factor this out 9x^2+6x+1=0? Icant seem to comprehend the steps

### 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 ,

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

Yes, that approach works perfectly.

Cheers,
Brent

### can you explain to me the one

can you explain to me the one that says quant a= n^2. I solved with two different sets of numbers, one gave me equal and one gave me not equal

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

It is the second question under difficulty level 150-159

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

I apologize, I wrote the 3 as a 2 by mistake.

### No problem. It happens to me

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

### Hey brent

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?

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?