If x + y = √3 and x - y = √2, then the value of 8xy(x2 + y2) is?
A. 6
B. √6
C. 5
D. √5
Answer: Option C
Solution(By Examveda Team)
$$\eqalign{ & x + y = \sqrt 3 \,.......{\text{(i)}} \cr & x - y = \sqrt 2 \,.......(ii) \cr & {\text{From equation (i) and (ii)}} \cr & x = \frac{{\sqrt 3 + \sqrt 2 }}{2} \cr & y = \frac{{\sqrt 3 - \sqrt 2 }}{2} \cr & {\text{So, }}8xy\left( {{x^2} + {y^2}} \right) \cr & = 8 \times \frac{{\sqrt 3 + \sqrt 2 }}{2} \times \frac{{\sqrt 3 - \sqrt 2 }}{2}\left[ {\frac{{{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}}}{4} + \frac{{{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}}}{4}} \right] \cr & = 2\left( {3 - 2} \right)\left[ {\frac{{3 + 2 + 2\sqrt 6 + 3 + 2 - 2\sqrt 6 }}{4}} \right] \cr & = 2 \times 1 \times \frac{{10}}{4} \cr & = 5 \cr} $$Join The Discussion
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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. $$\frac{8}{6}$$
x + y = √3
x – y = √2
∴ (x + y)2 + (x – y)2 = 3 + 2
⇒ 2 (x2 + y2) = 5 ...(i)
Again,
(x + y)2 – (x – y)2 = 3 – 2
⇒ 4xy = 1 ...(ii)
∴ (x2 + y2) = 5 × 1 = 5