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(?² − 5)⁴ + (?² − 7)⁴ = 16
(?² − 6 + 1)⁴ + (?² − 6 − 1)⁴ = 16
Take x² − 6 = t….(i)
(t + 1)⁴ + (t − 1)⁴ = 16
(t⁴ + 4 t³ + 6 t² + 4t + 1) + (t⁴ − 4t³ + 6t² − 4t + 1) = 16
2(t⁴ + 6t² + 1) = 16
t⁴ + 6t² − 7 = 0
Let t⁴ = k²
k² + 6k − 7 = 0
(k + 7)(k − 1) = 0
k = − 7, k = 1
k = t² = − 7 (not possible, because a perfect square can't be negative)
∴ t² = 1
t = ± 1
Putting t = 1 in the equation (i), we get
x² − 6 = 1
x² = 7
x = ± √7
Putting t = − 1 in the equation (i), we get
x² − 6 = − 1
x² = 5
x = ± √5
Difference between maximum root (+ √7) and minimumroot − √7) is
√7 − (− √7) = 2√7