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Mathematics

Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.

Quadratic Equations

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Answer

Let the larger number be x , so the smaller number is 16 - x.

Given, twice the square of the larger part exceeds the square of the smaller part by 164

∴2x2−(16−x)2=164⇒2x2−(256+x2−32x)=164⇒2x2−256−x2+32x−164=0⇒x2+32x−420=0⇒x2+42x−10x−420=0⇒x(x+42)−10(x+42)=0⇒(x−10)(x+42)=0⇒x−10=0 or x+42=0x=10 or x=−42\therefore 2x^2 - (16 - x)^2 = 164 \\[1em] \Rightarrow 2x^2 - (256 + x^2 - 32x) = 164 \\[1em] \Rightarrow 2x^2 - 256 - x^2 + 32x - 164 = 0 \\[1em] \Rightarrow x^2 + 32x - 420 = 0 \\[1em] \Rightarrow x^2 + 42x - 10x - 420 = 0 \\[1em] \Rightarrow x(x + 42) - 10(x + 42) = 0 \\[1em] \Rightarrow (x - 10)(x + 42) = 0 \\[1em] \Rightarrow x - 10 = 0 \text{ or } x + 42 = 0 \\[1em] x = 10 \text{ or } x = -42

Since, numbers are natural hence, x ≠ -42.

∴ x = 10 , 16 - x = 6.

Hence, the required numbers are 10, 6.

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