What is the value of t? (1) t(t - 1)^2 = 4t (2) t is an even integer

Question

What is the value of t?

(1) t(t - 1)^2 = 4t

(2) t is an even integer.

QuantMadeEasy · Answer

Step 1: Understanding statement 1 alone 
(1) t(t - 1)^2 = 4t
t(t - 1)^2 - 4t = 0
t {(t - 1)^2 - 4} = 0
t (t^2 + 1 - 2t -4) = 0
t (t^2 - 2t -3) = 0
t (t^2 -3t + t -3) = 0
t (t-3) (t+1) = 0
Hence, t=-1, 0 or 3
Insufficient

Step 2: Understanding statement 2 alone 
(2) t is an even integer.
t = 0, 2, 4....
Insufficient

Step 3: Combining statement 1 and 2 
t = 0
Sufficient

IMO C

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exc4libur · Answer

(1) insufic
t(t^2+1-2t)-4t=0, t^3+t-2t^2-4t=0,
t(t^2-2t-3)=0, t  =0,
t={3,0,-1}

(2) insufic

(1/2) sufic
t=0

(C)

IanStewart · Answer

Using Statement 1, t can clearly be 0. If t is not zero, we can divide by t on both sides to get (t-1)^2 = 4. If we square something and get 4 as our answer, we must be squaring either 2 or -2. So t-1 must either equal 2 or -2, and t itself must either equal 3 or -1. So we can't find a unique value for t using either Statement alone, but using both, t can only be zero, and the answer is C.

nehasomani33 · Answer

@thinkvision711

Can you please help!

shanky315 · Answer

Hi  Bunuel ,
Why is the answer E? using both the statement we can get t=0 (even number ). Please share your solution.

Thanks

Bunuel · Answer

The correct answer is C, not E. Edited. Thank you.

Kinshook · Answer

Asked: What is the value of t?

(1) t(t - 1)^2 = 4t
If t=3; 3*2^2 = 4*3; t=3 is a solution
t{(t^2 - 2t + 1-4} = 0
t(t^2 - 2t-3) = 0
t(t-3)(t+1) = 0
t = {0,3,-1}
NOT SUFFICIENT

(2) t is an even integer.
Multiple values of t are possible
NOT SUFFICIENT

(1) + (2)
(1) t(t - 1)^2 = 4t
(2) t is an even integer.
t = 0
SUFFICIENT

IMO C

Alankrita26 · Answer

But 0 doesn't come under even/odd category.

Bunuel · Answer

ZERO:

1. Zero is an  INTEGER .

2. Zero is an  EVEN  integer. An even number is an   integer   that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. Zero is  neither positive nor negative  (the only one of this kind).

4. Zero is  divisible by EVERY integer except 0 itself  ( \frac{x}{0} = 0 , so 0 is a divisible by every number, x).

5. Zero is a  multiple of EVERY integer  ( x*0 = 0 , so 0 is a multiple of any number, x).

6. Zero is  NOT a prime number  (neither is 1 by the way; the smallest prime number is 2).

7. Division by zero is NOT allowed:  anything/0 is undefined .

8. Any non-zero number  to the power of 0 equals 1  ( x^0 = 1 )

9.  0^0  case is  NOT tested on the GMAT .

10. If the exponent n is positive (n > 0),  0^n = 0 .

11. If the exponent n is negative (n < 0),  0^n  is undefined, because  0^{negative}=0^n=\frac{1}{0^{(-n)}} = \frac{1}{0} , which is undefined.  You CANNOT take 0 to the negative power .

12.  0! = 1! = 1 .

kelly_jacques · Answer

Bunuel

I understand this method, but I actually expanded the left side of the equation first, and left 4t on the right side. Thus, I had t^3 -2t^2 + t = 4t. Thereafter, I divided everything (both sides) by t. The problem with that method is that I do not get 0 as a root. I only end up with (t-3)(t+1) at the end. Could you please explain why my methodology is flawed and why we cannot divide by t?

Thank you.

Bunuel · Answer

You cannot divide t^3 - 2t^2 + t = 4t by t, or t(t - 1)^2 = 4t, specifically because t can be 0, and we cannot divide by 0. By doing so, you lose a root, namely t = 0.

Never divide an equation by a variable (or an expression containing a variable) unless you are certain that the variable (or the expression with the variable) does not equal zero. Division by zero is not allowed.

What is the value of t? (1) t(t - 1)^2 = 4t (2) t is an even integer

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What is the value of t? (1) t(t - 1)^2 = 4t (2) t is an even integer

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