Question : If $x^4+y^4+x^2 y^2=17 \frac{1}{16}$ and $x^2-x y+y^2=5 \frac{1}{4}$, then one of the values of $(x-y)$ is:
Option 1: $\frac{5}{2}$
Option 2: $\frac{3}{4}$
Option 3: $\frac{5}{4}$
Option 4: $\frac{3}{2}$
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Correct Answer: $\frac{5}{2}$
Solution :
Given: $x^4+y^4+x^2 y^2=17 \frac{1}{16}$..........................(1)
$x^2-x y+y^2=5 \frac{1}{4}$..........................(2)
⇒ $x^4 + y^4 + x^2y^2 = (x^2 - xy + y^2) (x^2 + xy + y^2)$
⇒ $17\frac{1}{16} = 5\frac{1}{4} (x^2 + xy + y^2)$
⇒ $\frac{273}{16} = \frac{21}{4} (x^2 + xy + y^2)$
⇒ $x^2 + xy + y^2 = \frac{13}{4}$..........................(3)
On adding (2) and (3), we get
$x^2 + y^2 = \frac{17}{4}$..................................(4)
Subtract (2) form (3), we get
$xy = -1$.....................................(5)
Now, Using the formula $(a - b)^2 = a^2 + b^2 - 2ab$
$(x - y)^2 = x^2 + y^2 - 2xy$
From (4) and (5), we get
⇒ $(x - y)^2 = \frac{17}{4} - 2 × (-1)$
⇒ $(x - y)^2 = \frac{17}{4} + 2 = \frac{25}{4}$
⇒ $(x - y) = \frac{5}{2}$
Hence, the correct answer is $\frac{5}{2}$.
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