Question

Can somebody please help me out with these 4 math problems?

Find the zeros of each quadratic function

a.) f(x)=x^2-16

b.)f(x)=x^2-7x+10

c.)f(x)=x^2-5x-24

d.) f(x)=2x^2-7x-4


(^2 means squared)

Answer 1

you can factor to linear terms to make it easy to solve when x=0
or just use the quadratic formula for x , plug in method for quadratic y=a*x^2+b*x+c = 0

Answer 2

There are different ways to go about solving these equations:
Method 1: Factorize.
Use these standard identities,
f(x)=x^2-a^2=(x+a)(x-a) which means that the zeroes are x=a or x=-a.
Example: f9x)=x^2-25=(x+5)(x-5) => x=5 or x=-5

There is a very detailed tutorial on factoring prepared by @bloomlocke367 here:
http://openstudy.com/study#/updates/55acffcee4b071e6530c96ce

Method 2: use the quadratic formula for the standard equation f(x)=ax^2+bx+c
\(\LARGE x=\frac{-b\pm\sqrt{b^2-4a}}{2a}\)
Zeroes of all of the above functions can be obtained using the formula.

Answer 3

* replace 4a by 4ac

Answer 4

so if i use method 2, would the answer i get be the zeros?

Answer 5

Yes.
For example, to solve
f(x)=x^2-11x+28
we have a=1, b=-11, c=28
substitute in formula:
\(\LARGE x=\frac{11\pm\sqrt{11^2-4(1)(28}}{2(1)}\)
\(\LARGE =\frac{11\pm\sqrt{121-112}}{2}\)
\(\LARGE =\frac{11\pm 3}{2}\)
\(=7~ or~ 4\)
so x=7 or x=4

Answer 6

Ohhh

Answer 7

Thank You

Answer 8

You're welcome!

Answer 9

Haha thanks @mathmate for the mention :D