If x4 + y4 + x2 + y2 = $$17\frac{1}{{16}}$$ and x2 - xy + y2 = $$5\frac{1}{4},$$ then one of the values of (x - y) is:
A. $$\frac{3}{4}$$
B. $$\frac{3}{2}$$
C. $$\frac{5}{4}$$
D. $$\frac{5}{2}$$
Answer: Option D
Solution(By Examveda Team)
$$\eqalign{ & {x^2} + {y^2} - xy = 5\frac{1}{4} \cr & {x^2} + {y^2} - xy = \frac{{21}}{4}........\left( {\text{i}} \right) \cr & {x^4} + {y^4} + {x^2}{y^2} = 17\frac{1}{{16}} \cr & {x^4} + {y^4} + {x^2}{y^2} = \frac{{273}}{{16}} \cr & {\left( {{x^2} + {y^2}} \right)^2} - {\left( {xy} \right)^2} = \frac{{273}}{{16}} \cr & \left( {{x^2} - xy + {y^2}} \right)\left( {{x^2} + xy + {y^2}} \right) = \frac{{273}}{{16}} \cr & \frac{{21}}{4}\left( {{x^2} + xy + {y^2}} \right) = \frac{{273}}{{16}} \cr & \left( {{x^2} + xy + {y^2}} \right) = \frac{{273}}{{16}} \times \frac{4}{{21}} \cr & \left( {{x^2} + xy + {y^2}} \right) = \frac{{13}}{4}........\left( {{\text{ii}}} \right) \cr & {\text{solve }}\left( {\text{i}} \right) + \left( {{\text{ii}}} \right) \cr & 2\left( {{x^2} + {y^2}} \right) = \frac{{21}}{4} + \frac{{13}}{4} \cr & 2\left( {{x^2} + {y^2}} \right) = \frac{{17}}{2} \cr & \left( {{x^2} + {y^2}} \right) = \frac{{17}}{4} \cr & {\text{solve }}\left( {\text{i}} \right) - \left( {{\text{ii}}} \right) \cr & - 2xy = \frac{{21}}{4} - \frac{{13}}{4} \cr & - 2xy = \frac{8}{4} \cr & - 2xy = 2 \cr & xy = - 1 \cr & {\left( {x - y} \right)^2} = {x^2} + {y^2} - 2xy \cr & {\left( {x - y} \right)^2} = \frac{{17}}{4} - 2 \times \left( { - 1} \right) \cr & {\left( {x - y} \right)^2} = \frac{{17}}{4} + 2 \cr & {\left( {x - y} \right)^2} = \frac{{25}}{4} \cr & x - y = \frac{5}{2} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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