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Bunuel
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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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
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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
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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)
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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
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