What is the value of x? 1) x^2 + x + 10 = 16 2) x = 4y^4+2y^2+2

Thanks for your explanation! I was thrown off by the extra variable "y" in the second statement, but when I got a positive and negative solution for x in statement one, that should've helped me figure out how to make use of the second.

Hi,

Can anyone clarify my query here?

Nothing is stated about 'y' rite...??
So y can be irrational, rational, integer, etc.

If i assume an extreme case: y an irrational number, then the answer is E rite?

Please clarify? Should i not take this case into consideration?

This is good question to help you make start thinking on a higher level for the GMAT and start for DS questions start asking the question "Why are they giving me this information?"

The question is x = ?

Statement 1:

When factored works out to be x = 2 or x = -3. This is clearly insufficient as we have two answers for x, one positive and one negative. When I solve and quadratic and I get one positive and one negative answer its usually a good idea to double check the question to make it doesn't provide a small bit of info that could eliminate one of the answer choices. In this case, there's no additional info so we'll carry on.

Statement 2:

Just by looking at the question we can see it's insufficient. Plug in 1 for y and we get one answer and plug in 2 and we get a different answer.

Statements 1 & 2:

This is where it might look insufficient. But if we look back at statement 2 and ask ourselves why did the test makers give us this info? We have one negative answer for x and one positive answer. Is there any value for y we can plug in that will give us an negative answer for x? No, all of the powers of Y are even so the answer will always be positive. Therefore, X = 2 and we pick answer C.

Target question: What is the value of x?

Statement 1: x² + x + 10 = 16
Given: x² + x + 10 = 16
Subtract 16 from both sides: x² + x - 6 = 0
Factor: (x + 3)(x - 2) = 0
So, EITHER x = -3 OR x = 2
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x = 4y⁴ + 2y² +2
Since we don't know anything about the value of y, there are many possible values of x. Here are two:
Case a: if y = 0, then x = 4(0)⁴ + 2(0)² +2 = 2
Case b: if y = 1, then x = 4(1)⁴ + 2(1)² +2 = 8
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that x = -3 OR x = 2
Statement 2 tells us that x is POSITIVE.
How do we know that x is POSITIVE?
Well, 4y⁴ ≥ 0 and 2y² ≥ 0 for all values of y. So, the SUM 4y⁴ + 2y² +2 MUST BE POSITIVE.
Since x = 4y⁴ + 2y² +2, we know that x is POSITIVE.
If x is positive, then x cannot equal -3, which means x must equal 2
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

(1) \(x^2 + x + 10 = 16\)
\(x^2 + x + 10 - 16 = 0\)
\(x^2 + x + 6 = 0\)
\(x^2 + 3x - 2x + 6 = 0\)
\(x(x + 3) - 2 (x + 3) = 0\)
\((x-2)(x + 3) =0\)

Therefore \(x\) could be \((2)\) or \((-3)\)

I is Not sufficient.

(2) \(x = 4y^4+2y^2+2\)

Value of \(x\) depends on value of \(y\). We do not know value of \(y\), hence we cannot find value of \(x\) from the equation alone.

II is Not Sufficient.

Combining (1) and (2)

From (1) we get \(x\) could be \((2)\) or \((-3)\)

From (2) we get \(x\) should be positive.

Hence \(x\) should be \(2\)

Answer (C)...

Solve that in a similar way. In condition 2 x must be even since all numbers on the right side are even. So, only possible answer is x = 2 considering condition 1.

You can only make that assumption if you know the numbers are integers. For instance, here's a much simpler problem:

x = 2y + 4

If you know that x and y are both integers, x definitely has to be even.

But if they don't have to be integers, you could say that y = 0.5. In that case, x = 1 + 4 = 5, which is odd.

Can someone please tell me if my approach is correct or incorrect:

If we plug in x as -3 into the second statement, we get 4y^4 + 2y^2 + 5 = 0 - but if we try to solve for a value of y, we are unable to because we cannot factor.

If we plug in x as 2 into the second statement, we get 4y^4 + 2y^2 = 0, we get either 2y^2 = 0, which means y must be 0, or we get 2y^2 = -1 which, when we simplify, get a negative root - y = root(-1/2). So only one case will work where y is equal to 0 -- so x must be 2? Is there anything wrong with this approach?

Thank you!

Hi Bunuel,

If you substitute X=-3 back into the equation, it does not satisfy L.H.S=R.H.S

Can you help me understand, when should we substitute back and when not? I am told that always to substitute back in such equations and in case of MODS.

?

I assume you are talking about the equation x^2 + x + 10 = 16 . If you substitute x=-3 in this, then LHS = (-3)^2 + (-3) + 10 = 9-3+10 = 16, which is same as RHS

But this same value of x=-3 cannot be substituted in second statement's equation (it wont make sense) x = 4y^4 + 2y^2 + 2 because RHS can never be negative, as other people (including Bunuel) have already explained in this thread.

From (1) x = -3 OR x = 2 and from (2) that x is a positive number, therefore when we combine the statements, x can only be 2. Sufficient.

Answer: C.[/quote]



Hi Bunuel,

If you substitute X=-3 back into the equation, it does not satisfy L.H.S=R.H.S

Can you help me understand, when should we substitute back and when not? I am told that always to substitute back in such equations and in case of MODS.

?[/quote]


I assume you are talking about the equation x^2 + x + 10 = 16 . If you substitute x=-3 in this, then LHS = (-3)^2 + (-3) + 10 = 9-3+10 = 16, which is same as RHS

But this same value of x=-3 cannot be substituted in second statement's equation (it wont make sense) x = 4y^4 + 2y^2 + 2 because RHS can never be negative, as other people (including Bunuel) have already explained in this thread.[/quote]


Apologies.. I made a stupid calculation mistake!!

(1) (x+3)(x-2) = 0, so x =-3 or x =2 NS

(2) x = 4y^4 + 2y^2 +2

ok, y =? NS

(1) and (2). At first glance it may seem there is not enough information here, however:

notice our variables are to even exponents on the RHS of statement 2. that means that the minimum value x can be is 2. sufficient

OA is C

(1) Solving the equation, we get x = -3, 2. INSUFFICIENT.

(2) Clearly insufficient -- we can get different values for y. INSUFFICIENT.

(1&2) Statement 1 tells us that x = -3 or 2. At first glance, the two statements combined might not seem sufficient. However, notice that y is raised to an even exponent, meaning that y is non negative.

Statement 2 actually tells us that x MUST be positive. We only have 1 positive solution, 2. SUFFICIENT.

Answer is C.

(1) Subtract 16 and factor : (x+3)(x-2) = 0 NS

(2) let a = y^2: 4a^2 +2a +2=0. Will have nonreal solutions NS

(1) and (2) We know (x+3)(x-2) = 0, so (4a^2+2a+2+3)=0----> non real solution or (4a^2+2a+2-2)=0 ---> 4a^2 +2a=0--> 2a(2a^2+1)=0, so 2a = 0 or (2a^2+1)=0 but nonreal sol, so 2a = 0 so a=0 so y^2=0 s0 x=0+0+2=2 sufficient

OA is E

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