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12 Days of Christmas GMAT Competition 2025 - Day 4: If x is a

1 2 3

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19 Dec 2025, 03:33

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Solving for x from the equation assuming the = sign and with respect to the non-negative condition we get x is not equal to 16
Looking at statement 1 it tells us that x is neither 1 nor 16 which isn't necessarily true because we only know x is not 16
Statement tells us x is neither 16 nor 18 which again cannot be inferred from the only condition that x is no 16
Statement 3 says is not among 4,9,16,25,36,49..... but we only know x is not equal to 16
Hence answer is none of the above E

Bunuel

If x is a non-negative number and $x ≠ 3\sqrt{x} + 4$, which of the following must be true?

I. (x - 1)(x - 16) ≠ 0

II. |17 - x| ≠ 1

III. x is not a square of an integer

A. I only
B. II only
C. III only
D. I, and II only
E. None of the above

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19 Dec 2025, 05:25

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Solving this question with reverse inequality to know which specific values of x are not allowed

x = 3 root x + 4

x-4 = 3 root x
(x-4)^2 = (3rootx)^2
Solve to get x=1 and x=16...(1)

St1: For this expression to be true, either x=1 or x=16

From (1), we know x is not =16, however if x=1, then (1-1)(1-16) is not = 0. St1 is not MBT.

St2: 17-x = 1 and 17-x = -1, x = 16 or x = 18. Since we already have x not = 16. This statement is also not always true.

St3: If x=1.3 or 4.5, does it violate condition in (1)? No.

None of the statements are always true. Option (E)

Bunuel

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19 Dec 2025, 05:50

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Bunuel

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Given $x ≠ 3\sqrt{x} + 4$

First let's find solution for the equality:
$(x - 4)^2 = (3\sqrt{x})^2 $

$ x^2 - 8x + 16 = 9x $

$ x^2 - 17x + 16 = 0 $

$ x^2 - 16x - x + 16 = 0 $

$ (x-16) (x-1) = 0 $

Now substituting in given inequality when x = 1, LHS = 1, RHS = 3 sqrt(1) + 4 = 7
So 1 cannot satisfy the given inequality

but x = 16 LHS = 16, RHS = 3*4 + 4 = 16
can satisfy the inequality

So solution is $ x ≠ 16 $

I. (x - 1)(x - 16) ≠ 0
x can be 1, so need not be true

II. |17 - x| ≠ 1

x can be 18 need not be true

III. x is not a square of an integer
x can be 1, 4, or any other perfect square except 16, need not be true.

None of the above.

Correct Answer: E

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19 Dec 2025, 05:50

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Let's square RHS and LHS.

(x-4)^2 != 9[sqrt(x)]^2
x^2 - 8x +16 != 9x
x^2 -17x +16 != 0
(x-1)(x-16) != 0 hence we have the I)

If you square the II) you get:
x^2 - 34x + 289 != 1 which does not always satistfy our initial equation, hence no II)

x is not the square of an integer?
again from the LHS and RHS we get x != 16 V x!=1 hence both squares of integers, hece we have the III)

Hence IMO E!

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19 Dec 2025, 06:20

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Let's solve the equation => x= 3\sqrt{}x +4

Let t= \sqrt{}x where t>=0
=> t^2=3t+4 => (t-4) (t+1) = 0.

Since, t>=0 => t=4 => x=16

So based the given condition x=16 (not equal to)

Testing each statement:

1. (x-1)(x-16)=0 => possible at x=1 => Not must be true
2. |17-x| = 1 => Not must be true at x=18
3. x is not a square of an interger. x=4 clearly works. Not must be true.

Hence none of the statements must be true => E

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19 Dec 2025, 06:46

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Find the forbidden values of x:
(x-4)^2 = 9x
x^2 - 8x + 16 = 9x
x^2 - 17x + 16 = 0
(x-1)(x-16) = 0
x=1 or x=16

But in the original expression only 16 is forbidden, 1 is allowed.

Forbidden value of x: 16

I. If x=1 (allowed value) then the expression is 0
II. If x=18 (allowed value) then the expression is 1
III. x can be, for example, 4 or 9

None must be true

IMO E

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19 Dec 2025, 07:01

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Given
1. x > 0
2. x != 3rootx + 4

Solving (2) we get
Let root(x) = y

y^2 = 3y + 4
=> y^2 -3y - 4 = 0
=> (y - 4)(y + 1)
=> y = 4 or y = -1

Since y = root(x) it will be positive => y = 4 => root(X) = 4 => x = 16

Hence the stem suggests that x != 16

So, conditions are x > 0 and x != 16

Evaluating the choices

1. (x-1)(x-16) != 0
x cannot be 16 but we don't know if x = 1, hence this cannot fulfil must be true criteria

2. |17 - x| != 1
Possible values of x are 18 and 16. Hence x !=16 and x != 18. We know about 16 but we don't know about 18 hence we cannot conclude that this must be true.

3. x is not a square of an integer
Values such 4, 9 are squares and fulfil the question stem. Again not 'must be true'

None of the above

Option E

Bunuel

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19 Dec 2025, 07:06

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If x is a non-negative number and x ≠ (3(x^0.5))+4, which of the following must be true?
First of all we need to find the values of x that are excluded.
Let, x= a^2
a^2 =3a+4
a^2 -3a -4=0
(a-4)(a+1) = 0
a=4 or a=-1 and x=a^2, hence x≠16 and x should be more than equal to 0.

I. (x - 1)(x - 16) ≠ 0
If x=1, then this statement is not true. Wrong
II. |17 - x| ≠ 1
If x=18, then this is not true. Wrong
III. x is not a square of an integer
Only the 16 is excluded but other values like (9,25,36...etc) are all allowed. Wrong

E. None of the above

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19 Dec 2025, 07:17

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Bunuel

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here 16 is the number for which equation fails

16 = 3* 4 + 4
16 = 16

Except 16, all number works

Let's check options
1. x is not equal to 16 & 1 , 16 is fine but x = 1 is fine, so it's partially true so it's no a must condition -- Not True

2. x can be 16 & 18, same logic as above x = 18 is fine --- Not true

3. we can check for 9 & 4 (as they are square of 2 & 3) and in both cases equation is true so it's not a must condition -- Not true

So our answer is E

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19 Dec 2025, 07:18

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t=sqrt(x) with t>=0

t^2=3t+4
t^2-3t-4=0
(t-4)(t+1)=0
t=4 or t=-1

but t>=0, so t=4 and x=16

I. This excludes x=1 and x=16, but the given condition only excludes x=16
II. This excludes x=16 and x=18, but the given condition only excludes x=16
III. x can be 1,4,9,25,36... so this is not necessarily true

Answer E

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19 Dec 2025, 07:23

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Bunuel

1st --> When x=16 ----> 3*4+4 = 16 hence, x must not be 16 .. must be true.

2nd ----> x = 16 or 18 ---> again x must not be 16 ...must be true.

3rd ----> let's try x = 64 ----> 3*8+4 = 28 not equal to 64... can be true

Hence, D is the answer

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19 Dec 2025, 07:40

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solving the inequality:
x-4 not equal to 3*sqrt(x)

squaring:
x^2-8x+16 not equal to 9x
x^2-17x+16 not equal to 0
(x-1)(x-16) not equal to 0

two solutions: 1 and 16

checking them in:
x-4 not equal to 3*sqrt(x)

only x=16 is not allowed

I. False if x=1
II. False if x=18
III. False if x=4

The answer is E

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I. If x=1 the expression is equal to 0 and 1 is not equal to 3*1+4=7
II. If x=18 the expression is equal to 1 and 18(integer) is not equal to 3*sqrt(18)+4(irrational)
III. If x=1(square of integer), 1 is not equal to 3*1+4=7

None of the above

The correct answer is E

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19 Dec 2025, 08:51

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!= means not equal to
>= means greater than or equal to
<= means lesser than or equal to

What we know is,
x is non negative
x != (3(x)^0.5) + 4 ------> Equation 1

Now lets say,
y = (x)^0.5
=> y^2 - 3y + 4 = 0
=> (y-4)(y+1) = 0

=> since y >= 0 , y!= 1
=> y = 4 and x = 16

=> Equation 1 => x!= 16

Now lets check each statement 1 by 1

Statement I. (x-1)(x-16) != 0
=> x!= 1 and x!= 16

We know x!=16, lets check for x!=1
Let, x = 1
=> x>=0
Substituting the value in equation 1
=> 1 != 7
But if we substitute 1 in (x-1)(x-16), we get 0 which cant be possible for I as it is given that it cant be equal to 0

Statement I not possible

Statement II =>. |17 - x | !=1

This is not possible if,
|17-x| = 1
=> x = 16 or x = 18

We know that x!= 16 , but we dont know about x = 18
We see that, for the given conditions,
18 >= 0
Substituting 18 in equation 1, we get
=> 18 != 3*((18)^(0.5)) + 4

=> 18 is allowed which causes the conditon II to fail

Statement II is not possible

Statement III - x is not a square of an integer

Here we need to check if any square other than 16 satisfies the given conditions

Lets say x = 9
=> 9 >= 0
=> 9 != 3*((9)^(0.5)) + 4 => 9 != 13 (Substitute 9 in equation 1)
=> Also 9 is a perfect square

We see that statement 3 is also not possible

E. None of the above

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19 Dec 2025, 10:45

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Bunuel, could you please with my below query :

Can we also solve like this :

x ≠ 3\sqrt{x} + 4[/m]
Therefore x-4 ≠ 3\sqrt{x} and subsequently square on both sides and solve till we get x not equal to 16 and x not equal to 1

Bunuel

Let's solve $x = 3\sqrt{x} + 4$.

Let $\sqrt{x}= t$, so $x = t^2$.

Substitute:

$t^2 = 3t + 4$
$t^2 - 3t - 4 = 0$

Factor:

$(t - 4)(t + 1) = 0$

So $t = 4$ or $t = -1$.

But $t = \sqrt{x}$ cannot be negative, so only $t = 4$ is valid.

Thus $\sqrt{x} = 4$, so $x = 16$.

Therefore $x ≠ 3\sqrt{x} + 4$ means $x ≠ 16$. That is the only restriction (along with x being a non-negative number).

Now test each statement to see whether it must be true when $x ≠ 16$.

I. (x - 1)(x - 16) ≠ 0

This expression is nonzero unless $x = 1$ or $x = 16$. We only know $x ≠ 16$, but x could be 1, which makes the expression 0. So Statement I is not guaranteed. Not always true.

II. |17 - x| ≠ 1

$|17 - x| = 1$ corresponds to $x = 16$ or $x = 18$. We only know $x ≠ 16$, but x could be 18, which would make $|17 - x| = 1$. So Statement II is not guaranteed. Not always true.

III. x is not a square of an integer

We only know $x ≠ 16$, but x could be 0, 1, 4, 9, 25, 36, etc. So Statement III is not guaranteed. Not always true.

Answer: E.

Bunuel

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19 Dec 2025, 10:51

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AditiDeokar

Squaring can create extra solutions, so anything you get after squaring must be checked in the original equation.

Here, squaring can produce x = 1, but x = 1 does not satisfy $x = 3\sqrt{x} + 4$, so it is extraneous. The only equality solution is x = 16, so the inequality just means x ≠ 16.

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Hey Bunuel,
Can you please confirm the solution as per the following logic :
I agree to all the above solutions where we are replacing x with square(t) but if we take a different route -
x-4 != 3*root(x)
and we square both sides, we get (x-4)^2 != 9x
now, when we simply the equation, we get :
x2 + 16 - 8x != 9x which becomes x2 - 17x + 16 != 0
on further simplification it becomes: (x-1)(x-16) != 0 which means x can't be 1 or 16 and hence option A in my opinion seems like a correct option
Can you please confirm why this logic is wrong ? What am I missing here ?