Question

If 4x^2+y^2=40 and xy=6. find the value of 2x+y and 2x-y.

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Answer 1

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Answer 2

GIVEN :-

• 4x² + y² = 40
• xy = 6

TO FIND :-

• value of 2x + y
• value of 2x - y

SOLUTION :-

for situation 1

$\implies \rm{2x + y}$

now squaring it

$\implies \rm{(2x + y) {}^{2} }$

we know the formula that

$\implies \boxed{\rm{(a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab} }$

$\implies \rm{(2x + y) {}^{2} = {(2x)}^{2} + {(y)}^{2} + 2(2xy) }$

$\implies \rm{(2x + y) {}^{2} = {4x}^{2} + {y}^{2} + 4(xy) }$

now we know the value that

4x² + y² = 40

xy = 6

hence putting the values

$\implies \rm{(2x + y) {}^{2} = 40+ 4(6) }$

$\implies \rm{(2x + y) {}^{2} = 40+ 24 }$

$\implies \rm{(2x + y) {}^{2} = 64 }$

$\implies \rm{(2x + y) = \sqrt{64} }$

$\implies \boxed{ \boxed{\rm{(2x + y) = 8 }}}$

for situation 2

$\implies \rm{2x - y }$

now squaring it

$\implies \rm{(2x - y) {}^{2} }$

we know the formula that

$\implies \boxed{\rm{(a - b) {}^{2} = {a}^{2} + {b}^{2} - 2ab} }$

$\implies \rm{(2x - y) {}^{2} = {(2x)}^{2} + {(y)}^{2} - 2(2xy) }$

$\implies \rm{(2x - y) {}^{2} = {4x}^{2} + {y}^{2} - 4(xy) }$

now we know the value that

4x² + y² = 40

xy = 6

hence putting the values

$\implies \rm{(2x - y) {}^{2} = 40 - 4(6) }$

$\implies \rm{(2x - y) {}^{2} = 40 - 24 }$

$\implies \rm{(2x - y) {}^{2} = 16 }$

$\implies \rm{(2x - y) = \sqrt{16} }$

$\implies \boxed{ \boxed{ \rm{(2x - y) = 4}}}$