Find the zeros of the following
USE ALL THE THREE METHODS FOR EACH NUMBER ( Factoring, Completing the Squares and Quadratic Formula)
1.f(x) =3x+5
2.g(y)=y^2-3y+2
3.s(xl= -6x^2-13x+5
Please help me I need this before 2 hours to come. Please help me hpw to do this .
Please help me how do i do the three methods for each number. I'll bring back the favor when time comes
first of all, the #1 is a linear function
So the three methods you are asked to use apply to a quadratic function.
So either silly or a typo.
I will do the hard one, you must be able to do the other.
-6x^2 - 13x + 5 = 0
6x^2 + 13x - 5 = 0
check if it factors:
6*(-5) = -30
does -30 have any factors that add up to -13, yes
so it does factor:
(3x - 1)(2x + 5) = 0
x = 1/3 or x = -5/2
by formula:
6x^2 + 13x - 5 = 0
x = (-13 ± √289)/12
= 1/3 or -5/2
completing the square on
6x^2 + 13x - 5 = 0
divide each term by 6
x^2 + (13/6)x + .... = 5/6 + ....
x^2 + (13/6)x + 169/144 = 5/6 + 169/144
assuming you know how to handle fractions ...
(x + 13/12)^2 = 289/144
x + 13/12 = ± 17/12
x = 17/12-13/12 or x = -17/12 - 13/12
x = 4/12 or x = -30/12
x = 1/3 or -5/2
On number 1, its really that form.. but i guess i cant use the factoring method there. So if i use the completing the squares and quadratic formula, will i add x^2 before the 3x+5?
( thank youfor helping reiny youre realy the best) ask me a favor next time . Love lots)
And in the number you answered did the dots mean
X^2+(13/6)x+13/12= -5/6 +13/12 ?
I remember my professor saying if gou add that 13 over 12 on the left side you should also add it on the right side but where the heck did the 13 or 12 came from ?
I really don't know why we add 13 over 12,
no, you can't do any of that.
Since the 3 methods apply to quadratics and you don't have a quadratic in #1, the question is bogus.
(That's like asking for 3 ways to bring a sailboat into a dock, when you sitting on a bicycle. )
re your question about the 13/12
in completing the square,
the first thing you do is divide each term by the coefficient of the square term so you have
x^2 in front
- then take half of the middle term coefficient:
1/2 of 13/6 is 13/12
now square that and add to both sides
(13/12)^2 = 169/144
and now
x^2 + (13/6)x + 169/144
is a perfect square, namely
(x + 13/12)^2
OH !!!!!! THANK YOUUU i now get it. Hahahah yey.
And heyyy reinyyy, they said u can do the number 1 either by completing the sqres or quad forkula.. if i use the quadratic formula will i let a be 0? And i ncompleting the squares wil i do a fornula to tet the first term?
No you can't use the formula for #1
If a = 0 , as it would be for the missing x^2 term, in the formula we are dividing by 2a
and since we can't divide by zero, any results would be totally meaningless.
So when "they" told you it could be done, tell "them" they are wrong.