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!!

4 Answers
Jun 12, 2018

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)

Jun 12, 2018

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)

Jun 12, 2018

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! :)

Jun 12, 2018

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"