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"#