Question

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

Answer 1

Hello from MrBillDoesMath!

Answer: 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

Answer 2

The quadratic equation is
`6y^2+4y+6=0` with
`y=(x-5)^2`.

Explanation:

A quadratic equation is written as:
`ax^2+bx+c=0, a\\eq 0`

We have the expression
`6(x-5)^4+4(x-5)^2+6=0` in order to form a quadratic equation, we can see that the both terms have in common
`(x-5)^2`.

Observation:
`((x-5)^2)^2=(x-5)^2^*^2=(x-5)^4`

Then we can substitute
`y=(x-5)^2`. And now we have to replace it.

`6(x-5)^4+4(x-5)^2+6=0\\6((x-5)^2)^2+4(x-5)^2+6=0\\6y^2+4y+6=0`

Then the quadratic equation is
`6y^2+4y+6=0` with
`y=(x-5)^2`.