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!!
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"