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

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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!!

This is what I got up to:

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!!

##### 4 Answers

The x intecepts are

#### Explanation:

As

or

#### 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 :

Then :

so

#### Explanation:

Solving for the x set

~Hope this helps! :)

#### 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"#