How many values of x are there for which 1/(x(x + 1)) +

Question

How many values of x are there for which \frac{1}{x(x+1)} + \frac{1}{x(x+2)} = \frac{1}{(x+1)} A. None B. One C. Two D. Three E. Four prep.png

gmatophobia · Accepted Answer

\frac{1}{x(x+1)} + \frac{1}{x(x+2)} - \frac{1}{x+1} = 0

Take x(x+1)(x+2) as the LCM

\frac{(x+2) + (x+1) -  }{x(x+1)(x+2)} = 0

\frac{x+2+x+1-x^2-2x}{x(x+1)(x+2)} = 0

\frac{-x^2 + 3}{x(x+1)(x+2)} = 0

Multiplying by -1

\frac{x^2 - 3}{x(x+1)(x+2)} = 0

The value is zero when  x^2 - 3 = 0

Hence,  x = \pm \sqrt{3}

Option C

chetan2u · Answer

The expression can be converted to quadratic equation so should have two values but let us check.

\frac{1}{x(x+1)}  +  \frac{1}{x(x+2)}  =  \frac{1}{(x+1)} 
Multiply by  x(x+1)(x+2) 
 x+2+x+1=x(x+2)…..2x+3=x^2+2x……x^2=3 
So x is  \sqrt{3}  or  -\sqrt{3}

C

Jayam12 · Answer

Hi gmatophobia

Needed your help in understanding this concept

https://www.manhattanprep.com/gmat/blog ... tive%20one.

So I read this article on how to deal with square roots in gmat. It says that we do not consider negative values after square root

Why is -(3)^1/2 correct?

Doesn't this question yield only one positive root?

Bunuel · Answer

Check the highlighted part below.

Mathematically,  \sqrt{...}  is the square root sign, a function (called the principal square root function), which cannot give negative result. So, this sign ( \sqrt{...} ) always means non-negative square root.

https://gmatclub.com/forum/download/file.php?id=39502 
 The graph of the function f(x) = √x

Notice that it's defined for non-negative numbers and is producing non-negative results.

TO SUMMARIZE:
 When the GMAT (and generally in math) provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root.  That is:

√9 = 3, NOT +3 or -3;
 \sqrt {16} = 2 , NOT +2 or -2;

Notice that in contrast, the equation x^2 = 9 has TWO solutions, +3 and -3. Because x^2 = 9 means that x =-√9 =-3 or x = √9 = 3.

Hope it helps.

Purnank · Answer

1/x * ( 1/(x+1)  +  1/(x+2) ) = 1/(x+1)
1/x+1  +  1/x+2 = x/x+1
Simplify,
(x+2+x+1) / (x+1)*(x+2)  = x / x+1
2x+3 / x+2 = x
2x+3 = x^2 + 2x
cancel out 2x
x^2 = 3
take square root 
2 values possible answer C.

Pr4n · Answer

If someone solved it the way I did, either they will cut two x's or they will not cut anything solve as it is in which case x=0 could also be a solution but notice that x can't be 0 as then 1/0 will be formed in the original eqn which mathematically isn't correct, hence only 2 solutions.

But if you cut x with each other it is correct for this question as it is an equality and not an inequality so you don't need to worry about the sign, second looking at the denominator we can tell x is not 0 so you can actually cancel the x's which each other and still get the right solution of 2 solutions.

I was confused between 2 & 3 when i face this question in a mock test, so putting this out there for anyone who chose 3 instead of 2, as they found x=0 also a solution when they solved the equation.

devdattakhoche · Answer

Bunuel ,  KarishmaB

Is there a faster way to solve this?
It took me ~3.5 minutes because at first glance it looks like I need to take the LCM and expand into a cubic. Only later I noticed the (x + 1) common factor, which reduces it to a quadratic.

But is there a shortcut or a recognition trick so that I don’t start expanding into a cubic in the first place? Any conceptual way to see the simplification earlier, instead of solving everything algebraically?

Looking for a more efficient GMAT-style approach.

Thanks!

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=86749 --- Harsha GMAT-Club-Forum-yexvi5bj.png

devdattakhoche · Answer

Good one, this makes more sense.

Thinking in direction that.. we see common terms in denominators , so just need to find suitable multiplication, this helps solve this questions really fast in like 20-25 seconds.

lawyeroffduty · Answer

can anyone share more such question of these types, it would be very helpful. thank you in advance.

Bunuel · Answer

Algebra

Theory  
 
  Math : Algebra 101  
  Algebra: Tips and hints  
  FOIL on the GMAT: Simplifying and Expanding  
  Factoring Quadratics  
  Solving Quadratic Equations   
  Using Algebra vs. Logic on GMAT Quant Questions

Questions  
 
  Bunuel's Algebra PS Diagnostic Test  
  Tough Algebra Set  
  Algebra Ability Quiz by e-GMAT  
  PS Questions

For more check  Ultimate GMAT Quantitative Megathread

Hope it helps.

DishaAgarwal12 · Answer

why can't x be 0,-1,-2

Bunuel · Answer

Because for x = 0, -1, or -2, the denominators become zero, so the equation is not defined. Remember, division by 0 is not allowed.

SwethaReddyL · Answer

Bunuel  /  KarishmaB  can you please tell me if the below approach is right?
I solved it this way,

i took the lcm after that,

x+2+x+1 (x+1) = x(x+1)(x+2)
2x+3(x+1) = x(x+1)(x+2)
5x+3 = 3 consecutive numbers

if i substitute values for x, the 3 consecutive numbers should be divisible by 3!, which is 6
by doing so, i get 2 values x = 5, 9
so i picked 2
but thinking of post 10, there are values divisible by 6 if substitute for x

help please

chetan2u · Answer

Two points..

1)
x+2+x+1 (x+1) = x(x+1)(x+2)

How did you simplify the LHS of equation to get 2x+3(x+1)

2x+3(x+1) = x(x+1)(x+2)

2) x can be any real number, need not be integers.
So x(x+1)(x+2) need not be product of three consecutive numbers..
It could be 2.5x3.5x4.5

egmat · Answer

Hi SwethaReddyL,

You landed on the right count ( 2 ), but the route you took isn't valid, and it's worth seeing why so it doesn't backfire on a similar question.

Let me start where your setup goes off track. Look at the clean version gmatophobia and chetan2u posted. Taking  x(x+1)(x+2)  as the common denominator:

- Left numerator: (x+ 2 ) + (x+ 1 ) =  2x+3 
- Right side, 1/(x+ 1 ), written over the same denominator, has numerator  x(x+2)

So the whole thing collapses to:

2x + 3 = x(x+2)  -  2 x+ 3  = x2+ 2 x -  x2 = 3  - x = ±√ 3 .

In your work you wrote  2 x+ 3 (x+ 1 ) = x(x+ 1 )(x+ 2 ) and kept the equation as a  cubic  (a product of three consecutive terms). That happened because the common  (x+1)  factor was never cancelled - it got dragged into both sides. Once you cancel it correctly, you're left with a simple quadratic, not a cubic.

The bigger issue is the divisibility step.

This equation never says x is an integer. There's no rule forcing x(x+ 1 )(x+ 2 ) to be " 3  consecutive numbers divisible by  6 " - that's a number-property fact about  integers , and it simply doesn't apply to an equation you're solving for  any  real x. The true solutions here,  ±√3 , aren't integers at all. Your x =  5  and x =  9  don't actually satisfy the original equation - check x= 5 : the left side is tiny fractions that can't equal  1/6 . Getting " two " was a coincidence.

The takeaway:  solve these by simplifying and taking roots, never by hunting for integers that make a product divisible by something. Equations give you roots through algebra - and those roots are allowed to be irrational.

Answer: C

SwethaReddyL · Answer

thanks much team. I get it now.
if i would've solved it like how you did, i am not sure if i would've included negative root. that is something i am still not clear about. isn't negative roots not considered in gmat?

KarishmaB · Answer

\frac{1}{x(x+1)} + \frac{1}{x(x+2)} = \frac{1}{(x+1)}

\frac{1}{(x+1)} + \frac{1}{(x+2)} = \frac{x}{(x+1)}

\frac{1}{(x+2)} = \frac{(x-1)}{(x+1)}

x + 1 = (x+2)(x-1)

x^2 = 3

x = \sqrt{3}  or  - \sqrt{3}

Answer (C)

Usernamevisible · Answer

Think:

"Multiply both sides by the LCM."

It takes about 20 seconds:

LCM: x(x+1)(x+2)
Cancel each denominator.
Solve the resulting quadratic.
Check excluded values.

This is the standard approach for rational equations.

How many values of x are there for which 1/(x(x + 1)) +

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How many values of x are there for which 1/(x(x + 1)) +

Prep Toolkit

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