if 4x^2 +y^2 =40 and xy=6 then find the value of 2xy +y
Answers
Answer:
The value of 2x+y is \bold{\pm 8}±8 .
Explanation:
Given:
4x^{2} + y^{2} = 404x
2
+y
2
=40
xy = 6xy=6
To find:
2x+y
Solution:
We know that,
(a+b)^{2} = a^{2} + 2ab + b^{2}(a+b)
2
=a
2
+2ab+b
2
According to the Given Problem,
4x^{2} + y^{2} = 40 \rightarrow(1)4x
2
+y
2
=40→(1)
xy=6 \rightarrow(2)xy=6→(2)
Square of (2x+y),
(2x+y)^{2} = (2x)^{2}+(2 \times 2 x \times y)+y^{2}(2x+y)
2
=(2x)
2
+(2×2x×y)+y
2
= 4x^{2}+4 x y+y^{2}=4x
2
+4xy+y
2
= \left(4 x^{2}+y^{2}\right)+4(x y)=(4x
2
+y
2
)+4(xy)
From 1st Equation and 2nd Equation:
=40+(4 \times 6)=40+(4×6)
(2 x + y)^{2} = 40 + 24(2x+y)
2
=40+24
(2 x + y)^{2} = 64(2x+y)
2
=64
Therefore,
2 x + y = \pm \sqrt{64}2x+y=±
64
2 x + y = \pm 82x+y=±8
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