If x + y = 6 and x² + y² = 20, find x⁴ + y⁴.
272
Given \(x + y = 6\) and \(x^2 + y^2 = 20\).
Squaring the first equation: \((x+y)^2 = x^2 + 2xy + y^2 = 36\).
So \(2xy = 36 - 20 = 16 \Rightarrow xy = 8\).
Now, \(x^4 + y^4 = (x^2+y^2)^2 - 2(xy)^2 = 20^2 - 2(8)^2 = 400 - 128 = 272\).
Hence, the value of \(x^4+y^4\) is 272.
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Important Questions from Identities
Simplify.
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- \(\dfrac{(0.73)^3+(0.31)^3}{(0.73)^2-0.73\times0.31+(0.31)^2}\)
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