DIVIDE 16 into two parts such that twice the square of larger part exceeds the square of the smaller part by 164.
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DIVIDE 16 into two parts such that twice the square of larger part exceeds the square of the smaller part by 164.
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according to question
2(x)sqr=(16-x)+164
2xsqr=256+xsqr-32x
xsqr+32x-256=0
factorise it
Let the two parts be x and 16 - x.
Given that Twice the square of larger part exceeds the square of the smaller part by 164.
= > 2x^2 = (16 - x)^2 + 164
= > 2x^2 - (16 - x)^2 = 164
= > 2x^2 - (256 + x^2 - 32x) = 164
= > 2x^2 - 256 - x^2 + 32x = 164
= > x^2 + 32x - 420 = 0
= > x^2 + 42x - 10x - 420 = 0
= > x(x + 42) - 10(x + 42) = 0
= > (x - 10)(x + 42) = 0
= > x = 10, x = -42
Therefore,
Larger part = 10.
Smaller part = 16 - 10
= 6.
Hope this helps!