Write 6(x – 5)4 + 4(x – 5)2 + 6 = 0 in the form of a quadratic by using substitution

Hello from MrBillDoesMath!

   6 u^2 + 4u + 6 = 0 where u = (x-5)^2

Discussion; I think the problem statement should actually be to rewrite this equation:

6 (x-5)^4 + 4 (x-5)^2 + 6 -= 0.

Note  the power of "x-5" is 4 in the first term and is 2 in the second term. That is, the power of x-5 in the first terms is double, or the square, of the (x-5) power occurring in the second term. This suggest the substitution u = (x-5)^2.  Then the equation can be rewritten as

6 u^2 + 4u + 6 = 0

which is a quadratic in "u".

Regards, MrB

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