Verified by Ajay Kumar
If x + y = 6 and x² + y² = 20, then find the value of (x⁴ + x²y² + y⁴).
412
336
408
372
Answer: B
Solution
Using \((x+y)^2=x^2+y^2+2xy\), \(36=20+2xy\), so \(xy=8\). Required value = \((x^2+y^2)^2-x^2y^2=20^2-8^2=336\).
यदि x + y = 6 और x² + y² = 20 है, तो (x⁴ + x²y² + y⁴) का मान ज्ञात करें।
412
336
408
372
उत्तर: B
हल
सूत्र \((x+y)^2=x^2+y^2+2xy\) से, \(36=20+2xy\), अतः \(xy=8\)। अभीष्ट मान = \((x^2+y^2)^2-x^2y^2=20^2-8^2=336\)।
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