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:
The quadratic equation is [tex]6y^2+4y+6=0[/tex] with [tex]y=(x-5)^2[/tex].
Step-by-step explanation:
A quadratic equation is written as: [tex]ax^2+bx+c=0, a\neq 0[/tex]
We have the expression [tex]6(x-5)^4+4(x-5)^2+6=0[/tex] in order to form a quadratic equation, we can see that the both terms have in common [tex](x-5)^2[/tex].
Observation: [tex]((x-5)^2)^2=(x-5)^2^*^2=(x-5)^4[/tex]
Then we can substitute [tex]y=(x-5)^2[/tex]. And now we have to replace it.
[tex]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[/tex]
Then the quadratic equation is [tex]6y^2+4y+6=0[/tex] with [tex]y=(x-5)^2[/tex].
