Question

How do you work out the x intercepts of the equation `y=2(x-4)^2-3` by factorising? Thanks!

This is what I got up to:
`y=2(x-4)^2-3`
`0=2(x-4)^2 - sqrt3^2`
`0=2(x-4 -sqrt3)(x-4+sqrt3)`
I know the third step is wrong because you have to include the 2 somewhere, but when I apply the null factor law the 2 sort of disappears and I get it wrong. Please help!!

Answer 1

The x intecepts are `x_1=4+sqrt(3/2)` and `x_2=4-sqrt(3/2)`

Explanation:

As `a^2-b^2=(a+b)(a-b)`, you could write the function as:
`y=[sqrt2(x-4)-sqrt3][sqrt2(x-4)+sqrt3]`

`y=0` whenever either of the parentheses is 0, i.e.
`sqrt2(x-4)-sqrt3]=0`, i.e `x_1=4+sqrt(3/2)`
or `sqrt2(x-4)+sqrt3]=0`, i.e. `x_2=4+sqrt(3/2)`

Answer 2

`x = 4+-sqrt(3/2)`

Explanation:

The solution is a lot simpler than you think. Just add a 3 to both sides of the equation and then you can get rid of the 2. Like this :

`0 = 2(x-4)^2 - 3`
`3 = 2(x-4)^2`
`3/2 = (x-4)^2`

Then :

`(x-4)^2 - sqrt(3/2)^2 = 0`
`(x-4 - sqrt(3/2))*(x-4 + sqrt(3/2)) = 0`

so `x = 4 + sqrt(3/2)` or `x =4 - sqrt(3/2)`

Answer 3

`color(red)(=> x=4+-sqrt(3/2)`

`color(magenta)(=>x=4+sqrt(3/2) or x=4-sqrt(3/2)`

Explanation:

`y=2(x-4)^2-3`

Solving for the x set `=>y=0`

`=>0=2(x-4)^2-3`

`=>2(x-4)^2=3`

`=>(x-4)^2=3/2`

`=>x-4=sqrt(3/2)`

`color(red)(=> x=4+-sqrt(3/2)`

`color(magenta)(=>x=4+sqrt(3/2) or x=4-sqrt(3/2)`

~Hope this helps! :)

Answer 4

`x=4+-sqrt(3/2)`

Explanation:

`"to solve for x set "y=0`

`2(x-4)^2-3=0`

`"add 3 to both sides"`

`2(x-4)^2=3`

`"divide both sides by 2"`

`(x-4)^2=3/2`

`color(blue)"take the square root of both sides"`

`sqrt((x-4)^2)=+-sqrt(3/2)larrcolor(blue)"note plus or minus"`

`x-4=+-sqrt(3/2)`

`"add 4 to both sides"`

`x=4+-sqrt(3/2)larrcolor(red)"exact values"`

`x=4+sqrt(3/2)~~5.22" to 2 dec. places"`

`"or "x=4-sqrt(3/2)~~2.78" to 2 dec. places"`