Simplifying Quadratic Equations

May be Useful for you guys...
FINDING THE COMMON ROOTS
Let f(x) = 0 and g(x) = 0 be two quadratic equations in x then their common roots will also be the roots of the equation f(x) – g(x) = 0.
The roots of f(x) – g(x) = 0 may or may not be the roots of f(x) = 0 and g(x) = 0. Therefore, to find the common roots of f(x) = 0 and g(x) = 0 follow the following procedure.
Step 1:
Find f(x) – g(x)
Step 2:
Find the root(s) of f(x) – g(x) = 0
Step 3:
Substitute the roots of f(x) – g(x) = 0 in f(x) and g(x). If the roots satisfy both the equations then they are the common roots.
Example 5: The number of roots common between the two equations x3 + 3x2 + 4x + 5 = 0 and x3 + 2x2 + 7x + 3 = 0 is:
Solution: x3 + 3x2 + 4x + 5 = 0... (i)
x3 + 2x2 + 7x + 3 = 0... (ii)
Equating equations (i) and (ii), we get,
x2 – 3x + 2 = 0
(x – 1)(x – 2) = 0
x = 1 or x = 2
Since both x = 1 and x = 2, do not satisfy either (i) or (ii), there exists no common roots for equations (i) and (ii)

Find more useful tips and tricks here - https://aptidude.com/trickbook

Hello Rich,

Thanks for your explanation.
However, I cannot get this part :

We'll start by factoring out (2T-1) from the numerator....
[(2T-1)(1 + 2T-1)]/(2T-1)

(1 + 2T - 1) --> Where does the expression come from ?

I would solve like :
[(2T-1) + (2T-1)^2]/(2T-1)
(2T - 1)/(2T- 1) = 1
(2T-1)^2/ (2T-1) =(2T-1)*(2T-1)/ (2T-1) = 2T-1

Result 1+2T-1 = 2T

Am I right?

Thanks a lot,
Daniela

Hi dmatinho,

The factoring involved in this question is a bit more complicated than normal since we're factoring out a "term" and not just an individual number or variable.

For example, factoring the following is pretty easy:

2X - 4

2(X - 2)

Since you can "see" that there's a "2" in both terms

In this prompt, we have to factor out (2T-1) from each of the terms:

[(2T-1) + (2T-1)^2]/(2T-1)

I'm going to split the numerator into 2 'pieces'....

(2T-1) and (2T-1)^2

Now, think about how you would factor (2T-1) out of each piece...

(1)(2T-1) and (2T-1)(2T-1)

Next, when we completely factor out (2T-1), think about what's "left over"....

(2T-1)[ (1) + (2T-1)]

Inside the bracket, the +1 and the -1 'cancel out', leaving us with...

(2T-1)(2T)

The last step is to deal with the denominator of the original fraction....

(2T-1)(2T)/(2T-1)

That denominator cancels out the same term in the numerator, leaving us with....

2T

The approach that you took is fine; you're ultimately just breaking the fraction into two pieces and then combining the results later.

GMAT assassins aren't born, they're made,
Rich

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.