Question

PLZZZ Help!!! The question is " find the zeros of each function by using the quadratic formula."

h(x) = 3x^2-3x+3/4

Answer 1

`(1)/(2)`

Explanation:

Quadratic formula goes as follows
`(-b+√(b^2-4ac) )/(2a)` and
`(-b-√(b^2-4ac) )/(2a)`, there is no plus or minus (+/- thingy) sign on here but use both plus and minus on the numerator where I switched the signs.

The equation you give is
`3x^2-3x+(3)/(4)`, to make it nice, multiply everything by 4 to cancel out the fraction and end up with
`12x^2-12x+3`. So a=12, b=-12, and c=3. Use these values and plug them into quadratic formula.

For the first formula I wrote,
`(-b+√(b^2-4ac) )/(2a)`, you get
`(-(-12)+√((-12)^2-4(12)(3)) )/(2(12))`=
`(1)/(2)`.

For the second formula,
`(-b-√(b^2-4ac) )/(2a)`, you get
`(-(-12)-√((-12)^2-4(12)(3)) )/(2(12))`=
`(1)/(2)`.

Always do quadratic formula for both plus and minus to get both zeros. However, in this case, the answer still remained positive. So the zero of this would be just
`(1)/(2)`.