Question

(√3+√2)⁴+(√3-√2)⁴=98​

Answer 1

Let f(x) = (x+3)^4 + (x+2)^4 + (x+1)^4

As other answerers have done, make the substitution t = x+2 so that

f(t) = (t+1)^4 + t^4 + (t-1)^4

After solving for f(t) = 98 we can just subtract 2 from the roots to get the roots for f(x) = 98

Now note that f(t) is in fact an even function (symmetric about the y-axis). This means in particular if a is a root, then so is -a. Since f(t) is a 4th degree polynomial, we may assume that f(t) = 98 has the roots +/-a, and +/- b, i.e. we can express f(t)-98 as c(t+a)(t-a)(t-b)(t+b) = c(t^2-a^2)(t^2-b^2). Comparing coefficients of f(t)-98 and c(t^2-a^2)(t^2-b^2) for

t^4: c = 3

t^2: c(-a^2 -b^2) = 12 => a^2 + b^2 = -4

t^0: c(a^2b^2) = -96 => a^2b^2 = -32

So we can solve for t^2 via the quadratic equation (t^2)^2 + 4(t^2) - 32 = 0, giving t^2 = 4 or -8 => t = +/-2 or +/-2sqrt(2)i. Hence x = t-2 = 0, -4 or -2 +/-2sq

Answer 2

$\huge\bold\blue{hey friend}$

Answer:

98

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