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Mathematics

Find three succcessive even natural numbers, the sum of whose squares is 308.

Quadratic Equations

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Answer

Let the required numbers be x, x + 2, x + 4.

Given, the sum of squares of these numbers = 308

⇒x2+(x+2)2+(x+4)2=308⇒x2+x2+4+4x+x2+16+8x=308⇒3x2+12x+20=308⇒3x2+12x−288=0⇒3(x2+4x−96)=0⇒x2+4x−96=0⇒x2+12x−8x−96=0⇒x(x+12)−8(x+12)=0⇒(x−8)(x+12)=0⇒x−8=0 or x+12=0x=8 or x=−12\Rightarrow x^2 + (x + 2)^2 + (x + 4)^2 = 308 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x + x^2 + 16 + 8x = 308 \\[1em] \Rightarrow 3x^2 + 12x + 20 = 308 \\[1em] \Rightarrow 3x^2 + 12x - 288 = 0 \\[1em] \Rightarrow 3(x^2 + 4x - 96) = 0 \\[1em] \Rightarrow x^2 + 4x - 96 = 0 \\[1em] \Rightarrow x^2 + 12x - 8x - 96 = 0 \\[1em] \Rightarrow x(x + 12) - 8(x + 12) = 0 \\[1em] \Rightarrow (x - 8)(x + 12) = 0 \\[1em] \Rightarrow x - 8 = 0 \text{ or } x + 12 = 0 \\[1em] x = 8 \text{ or } x = -12

Since the numbers are natural hence , x ≠ -12

∴ x = 8 , x + 2 = 10 , x + 4 = 12.

Hence , the required numbers are 8, 10, 12.

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