Question

f(x) = 1/2x^2 +5/2x−3/2

(a) Find all the real zeros of the polynomial function.
x = (smaller value)
x = (larger value)

Answer 1

am i supposed to start off by isolating x^2?

Answer 2

looks like you have to use the quadratic formula

set f(x)= 0
1/2x^2 +5/2x−3/2 = 0

(I assume this means (1/2) * x^2 (and not 1 over (2x^2))
multiply both sides of the equation by 2 (this simplifies it a little)
x^2 + 5x -3 =0

It does not factor nicely, so we use
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]
where a,b,c match the coefficients:
\[ ax^2 + bx +c =0\]

Answer 3

i have a question. if this was a simple (x+2)^2 type of thing, i wouldn't use that formula. only use that formula when it doesn't factor nicely?

Answer 4

If you can factor it easily. But the formula works: (x+2)^2=0 is
x^2 +4x +4 = 0
a=1, b=4, c=4 so
(-4 ±sqrt(16-4*1*4))/2 = (-4 ±0)/2 = -2
so x= -2
or x+2= 0
here the two solutions are both -2

Answer 5

i got -5+-√37)/2

Answer 6

thanks for your help!

Answer 7

yes , but don't forget the parens
(-5+-√37)/2 (2 divides both the -5 and the root)