divide 12 into two parts such that sum of their square is 79
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the two equal halves are under root (79/2)
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x+y = 12.
y=12-x. (1)
x^2 + y^2=79. (2)
now from eq 1 put value of y in eq 2
we get. x^2 + (12-x) ^2 = 79
x^2 + 144 + x^2 - 24x =79
2x^2 -24x + 144 = 79
2x^2 - 24x +144-79 = 0
2x^2 - 24x +64 =0
2x^2 - 16x - 8x + 64 = 0
2x(x-8) - 8(x-8) =0
(2x-8) (x-8) =0
2x-8=0. x-8=0
x=4. x= 8
now put value of x in eq 1 we get,
when, x=4. x=8
y = 12 - 4. y=12-8
y=8. y = 4
as we can see the two no. we get are 4 and 8
and these are the final answer of this question.
HOPE THIS WILL HELP YOU
y=12-x. (1)
x^2 + y^2=79. (2)
now from eq 1 put value of y in eq 2
we get. x^2 + (12-x) ^2 = 79
x^2 + 144 + x^2 - 24x =79
2x^2 -24x + 144 = 79
2x^2 - 24x +144-79 = 0
2x^2 - 24x +64 =0
2x^2 - 16x - 8x + 64 = 0
2x(x-8) - 8(x-8) =0
(2x-8) (x-8) =0
2x-8=0. x-8=0
x=4. x= 8
now put value of x in eq 1 we get,
when, x=4. x=8
y = 12 - 4. y=12-8
y=8. y = 4
as we can see the two no. we get are 4 and 8
and these are the final answer of this question.
HOPE THIS WILL HELP YOU
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