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Simplifying Quadratic Equations 26 Mar 2015, 13:58
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This question appears in the MGMAT foundations of math guide. Chapter 7 Review Questions, Drill 3, #19. I am having trouble figuring out the correct steps to take in simplifying (Even after following MGMATS steps)

19) simplify the equation: (if t is not equal to 1/2),

2t-1 + (2t-1)^2/2t-1 = ?

A) 2t

b) 2t-1

c) 2t-2

The correct answer is A, but would really appreciate a step by step 'how to' on how to get there. Thanks in advance for any help.

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Simplifying Quadratic Equations 26 Mar 2015, 18:41
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Hi paul.jepsen,

I think that part of what might be confusing you about this prompt is that the "common term" is (2T-1). Let's start with a simpler example (that's based on the exact same concept):

[X - X^2]/X

In the numerator, you can 'factor out' an X, which will give you....

[X(1 - X)]/X

Now you can divide the X "out" of the equation and you'll be left with....

(1-X)

We can do the exact SAME thing with fraction in this question:

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

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

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

Now we can divide the (2T-1) out of the equation and you'll be left with...

(1 + 2T-1)

Combine 'like' terms and you get...

2T

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Simplifying Quadratic Equations 27 Mar 2015, 11:15
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Thanks Rich. Very helpful. I was looking at the problem the wrong way, you cleared it up for me.
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Simplifying Quadratic Equations 02 Apr 2015, 00:20
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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
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Simplifying Quadratic Equations 18 Apr 2015, 02:30
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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

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Hi paul.jepsen,

I think that part of what might be confusing you about this prompt is that the "common term" is (2T-1). Let's start with a simpler example (that's based on the exact same concept):

[X - X^2]/X

In the numerator, you can 'factor out' an X, which will give you....

[X(1 - X)]/X

Now you can divide the X "out" of the equation and you'll be left with....

(1-X)

We can do the exact SAME thing with fraction in this question:

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

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

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

Now we can divide the (2T-1) out of the equation and you'll be left with...

(1 + 2T-1)

Combine 'like' terms and you get...

2T

GMAT assassins aren't born, they're made,
Rich
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Simplifying Quadratic Equations 18 Apr 2015, 10:31
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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

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