Question

How do you find the zeros of `f(x)=5x^2-25x+30`?

Answer 1

See a solution process below:

Explanation:

We can factor this function as:

`f(x) = (5x - 15)(x - 2)`

To find the zeros we can solve each term on the right side of the function for `0`:

Solution 1:

`5x - 15 = 0`

`5x - 15 + color(red)(15) = 0 + color(red)(15)`

`5x - 0 = 15`

`5x = 15`

`(5x)/color(red)(5) = 15/color(red)(5)`

`(color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5)) = 3`

`x = 3`

Solution 2:

`x - 2 = 0`

`x - 2 + color(red)(2) = 0 + color(red)(2)`

`x - 0 = 2`

`x = 2`

The Solutions Are: `x = {2, 3}`

Answer 2

`x=2,x=3`

Explanation:

`"to calculate the zeros set "f(x)=0`

`rArr5x^2-25x+30=0larrcolor(blue)"factorise to solve"`

`rArr5(x^2-5x+6)=0`

`"the factors of + 6 which sum to - 5 are -2 and - 3"`

`rArr5(x-2)(x-3)=0`

`"equate each factor to zero and solve for x"`

`x-2=0rArrx=2`

`x-3=0rArrx=3`

`"the zeros are "x=2,x=3`
graph{(y-x^2+5x-6)((x-2)^2+y^2-0.07)((x-3)^2+y^2-0.07)=0 [-10, 10, -5, 5]}