Answer:
Step-by-step explanation:
a) cannot -3x^2 and 3x^2 cancel out. What is left is a linear equation.
9x = 14 +x - 1
b) cannot use the quadratic formula to solve this equation the x on the right must be made positive. When it is, it is in the denominator as in [tex]\frac{3}{x^2}[/tex]When that happens, there is no way to get the 5x to remain as it is. One way or another you will wind up with 5x^3 + 4x^2 = 3 which is a cubic, and not a quadratic.
c) If c ends where I think it does, the quadratic will work on it.
-x^2 + 4x + 7 = 2x^2 - 9
0 = 2x^2 + x^2 - 4x - 7 - 9
0 = 3x^2 - 4x - 16
a = 3 ; b = - 4 and c = - 16
The solutions are (3.07,0) and (-1.74,0) both are rounded.
d)x^3 + x^2 + x = 0 could have the quadratic used on it. a = 1, b=1, c = 1 which a common factor of x is taken out by the distributive property.
x(x^2 + x + 1) = 0 I think the letter d for the question number has been omitted. The solution sets are (0,0) and 2 complex roots. But that was not what you were asking. You were just asking if you could use the formula and you can if I've separated out things correctly. If I have not then c cannot be solved with the quadratic.
Answer:
it's C and B, it is atleast
Step-by-step explanation:
as of March 2021
