Question

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

Answer 1

Answer:

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

Step-by-step explanation:

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

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 $y=(x-5)^2$.

Answer 2

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