Lesson: Basic Equation Solving

Comment on Basic Equation Solving

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

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)


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
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

greenlight-admin's picture

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

Yes, that approach works perfectly.


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
greenlight-admin's picture

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 under difficulty level 150-159
greenlight-admin's picture

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 a 2 by mistake.
greenlight-admin's picture

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?
greenlight-admin's picture

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?

Have a question about this video?

Post your question in the Comment section below, and a GRE expert will answer it as fast as humanly possible.

How do the Reinforcement Activities work?

Let me Know

Have a suggestion to make the course even better? Email us today!

Free “Question of the Day” emails!