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
If x 2 + 4y 2 = 40, xy = 6 and x > 2y then the value of x - 2y is:
- \(\dfrac{(5.17-2.19)^2-(5.17+2.19)^2}{11.3223}\)
If a(a + b + c) 2= 1792; b(a + b + c) 2= 1536; c(a + b + c) 2= 768 then what will be the value of a?
If a + b + c = 5, a 2+ b 2+ c 2= 27, a 3 + b 3 + c 3 = 125, then the value of \(\frac{abc}{5}\) is:
(999) 2- 2 2=
