If (x+y+z)=23 and x^2+y^2+z^2=179 , then find the value of (xy+yz+zx) . - Quantitative Aptitude MCQ SSC CGL 2026 | ExamTest.live

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If (x + y + z) = 23 and x^{2} + y^{2} + z^{2} = 179, then find the value of (x y + yz + z x) .

Solution & Step-by-step Explanation

We use the algebraic identity: (x + y + z)^{2} = x^{2} + y^{2} + z^{2} + 2 (x y + yz + z x) Substitute the given values into the equation: (23)^{2} = 179 + 2 (x y + yz + z x) 529 = 179 + 2 (x y + yz + z x) 2 (x y + yz + z x) = 529 - 179 2 (x y + yz + z x) = 350 x y + yz + z x = \frac{350}{2} = 175

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If (x + y + z) = 23 and x^{2} + y^{2} + z^{2} = 179, then find the value of (x y + yz + z x) .

280

225

175

350

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