If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what

Question

If the sum of the squares of x and y is 3, and  x^4 = y^4 + 25 , what is the value of  x^2 ?

A.  \frac{-8}{3}

B.  \frac{14}{3}

C.  \frac{17}{3}

D.  \frac{28}{3}

E.  \frac{34}{3}

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hr1212 · Accepted Answer

x^2 + y^2 = 3  --- (1)

x^4 - y^4 = 25

(x^2 - y^2)*(x^2 + y^2) = 25 
 (x^2 - y^2) = 25/3  --- (2)

Adding (1) and (2)

2x^2 = 3 + 25/3 
 2x^2 = 34/3  
 x^2 = 17/3

IMO: C

NextstopISB · Answer

x^2 + y^2 = 3
x^2 = y^2 + 25

Substitute x^2 = y^2 + 25 into x^2 + y^2 = 3:
(y^2 + 25) + y^2 = 3
2y^2 + 25 = 3
2y^2 = 3 - 25
2y^2 = -22
y^2 = -22 / 2
y^2 = -11

Now, substitute y^2 = -11 into x^2 = y^2 + 25:
x^2 = (-11) + 25
x^2 = 14

ADisHere · Answer

https://gmatclub.com/forum/download/file.php?id=83469 GMAT-Club-Forum-jaxi2yr4.jpeg

Gmat750aspirant · Answer

1) x^2 + y^2 = 3

2) x^4 - y^4 = 25

Rewriting equation (2):
(x^2 - y^2)(x^2 + y^2) = 25

Substitute x^2 + y^2 = 3 from equation (1):
(x^2 - y^2) * 3 = 25

So:
x^2 - y^2 = 25 / 3 (equation 2)

Adding equations (1) and (2):
(x^2 + y^2) + (x^2 - y^2) = 3 + 25 / 3

Simplifying:
2x^2 = 34 / 3

Finally:
x^2 = 17 / 3

Whoisdmx15 · Answer

Given  x^2+y^2 = 3  ...(i)
and  x^4 = y^4+25 ...(ii)

This gives us  x^4-y^4=25 
 (x^2+y^2)*(x^2-y^2) =25

Substituting from (i)
 3*(x^2-y^2) = 25 
 (x^2-y^2) = \frac{25}{3} ...(iii)

(i)+(ii)
 2*x^2=3+\frac{25}{3} 
Hence  x^2= \frac{17}{3}

IMO Ans C

eega · Answer

Given x^2+y^2 = 3......(1);
x^4= y^4 + 25.....(2)
x^2 = ?

x^2=3-y^2
sub it in (2)

(3-y^2)^2 = y^4 + 25

solving we get y^2 = -16/6
sub it in (1)
we get x^2 = 17/3  (Option D)

luisdicampo · Answer

Deconstructing the Question

We are given

x^2 + y^2 = 3

x^4 = y^4 + 25

We need to find   x^2  .

Step-by-step

Rewrite:

x^4 - y^4 = 25

Factor:

(x^2 - y^2)(x^2 + y^2) = 25

Substitute   x^2 + y^2 = 3  :

(x^2 - y^2)\cdot 3 = 25

x^2 - y^2 = \frac{25}{3}

Now add the equations:

2x^2 = 3 + \frac{25}{3} = \frac{34}{3}

x^2 = \frac{17}{3}

Answer: C

FerociousStorm · Answer

As shown by other solutions x^2 - y^2 = 25/3 so x^2 must be greater than 25/3 which is not 17/3. where am i going wrong ?

egmat · Answer

Hi FerociousStorm,

Your algebra is fine; the slip is one hidden assumption baked into "x2 must be greater than 25/3."

That statement is only true  if y2 is positive . You're reading x2 - y2 = 25/3 as "x2 = 25/3 + (something positive)," so x2 has to be bigger. But look at what y2 actually is here.

Find y2 directly: 
- From x2 + y2 =  3 , we get y2 = 3 - x2.
- With x2 =  17/3 , that gives y2 = 3 - 17/3 =  -8/3 .

So in this problem y2 is  negative  (that's actually the trap value in choice A). When y2 is negative, x2 = 25/3 + y2 = 25/3 - 8/3 =  17/3 , which is  less  than 25/3 - exactly what you found. No contradiction.

Here's the tell you can catch faster: x2 + y2 =  3  means that  if y2 were positive, x2 couldn't even exceed 3.  But the answer 17/3 ≈  5.7  is already bigger than  3 . That alone signals y2 must be negative. This is a pure algebra puzzle - the two equations together simply have no real y, and that's allowed unless the question says x and y are real.

Quick parallel to lock it in.  Solve this tiny system the same way:
- a2 + b2 =  1 
- a2 - b2 =  5

Add them: 2a2 =  6 , so a2 =  3  - even though a2 - b2 =  5  is bigger. Back-substitute: b2 = 1 - 3 =  -2 , negative. Same mechanism: subtracting a  negative  square makes the smaller-looking result perfectly consistent.

So your instinct "x2 > 25/3" quietly assumed y2 ≥  0  - drop that assumption and everything lines up to  C .

Answer: C

MaTaM · Answer

x^2+y^2=3  which means  y^2=3-x^2

x^4=y^4+25=(y^2)^2+25 
 x^4=(3-x^2)^2+25 
 x^4=9+x^4-6x^2+25 
 6x^2=34 
 x^2=34/6=17/3

If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what

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If the sum of the squares of x and y is 3, and x^4 = y^4 + 25, what

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