Answer:
- x^4 – 6x^2 – 27 = 0
- 6(2x + 4)^2 = (2x + 4) + 2
- 6x^4 = -x^2 + 5
Explanation:
You want to identify the equations that are quadratic in form.
Quadratic
A polynomial equation is "quadratic in form" if the powers of the variables are integer multiples of {0, 1, 2}, or some subset of these that includes 2. That is, it will be "quadratic" if some substitution z = f(x) is possible such that the resulting equation in z is a polynomial equation of degree 2.
x^4 – 6x^2 – 27 = 0
Term are degrees 0, 2, 4, so this is "quadratic in form". The substitution z=x^2 will give a quadratic in z.
3x^4 = 2x
Term degrees are 1 and 4, so this is not quadratic.
2(x + 5)^4 + 2x^2 + 5 = 0
The substitution z = (x+5)^2 is suggested by the 4th-power term, but the remaining terms cannot be written as a linear function of z. This is not quadratic.
6(2x + 4)^2 = (2x + 4) + 2
This is a straight quadratic equation. One could substitute z=2x+4, but that is not necessary. This is "quadratic in form."
6x^4 = -x^2 + 5
Term degrees are 0, 2, 4, so this is "quadratic in form."
8x^4 + 2x^2 – 4x = 0
Term degrees are 1, 2, 4, so this is not quadratic.