After being rearranged and simplified, which of the following equations could be solved using the quadratic formula? check all that apply. a) -3x^2+9x+3x^2=14+x-1 b)5x+4=3x^-2 c)-x^2+4x+7=2x^2-9 x^3+x^2+x=0

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 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.

its an acute right triangle

step-by-step explanation: