How do you work out the x intercepts of the equation y=2(x-4)^2-3y=2(x−4)2−3 by factorising? Thanks!
This is what I got up to:
y=2(x-4)^2-3y=2(x−4)2−3
0=2(x-4)^2 - sqrt3^20=2(x−4)2−√32
0=2(x-4 -sqrt3)(x-4+sqrt3)0=2(x−4−√3)(x−4+√3)
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=0to solve for x set y=0
2(x-4)^2-3=02(x−4)2−3=0
"add 3 to both sides"add 3 to both sides
2(x-4)^2=32(x−4)2=3
"divide both sides by 2"divide both sides by 2
(x-4)^2=3/2(x−4)2=32
color(blue)"take the square root of both sides"take the square root of both sides
sqrt((x-4)^2)=+-sqrt(3/2)larrcolor(blue)"note plus or minus"√(x−4)2=±√32←note plus or minus
x-4=+-sqrt(3/2)x−4=±√32
"add 4 to both sides"add 4 to both sides
x=4+-sqrt(3/2)larrcolor(red)"exact values"x=4±√32←exact values
x=4+sqrt(3/2)~~5.22" to 2 dec. places"x=4+√32≈5.22 to 2 dec. places
"or "x=4-sqrt(3/2)~~2.78" to 2 dec. places"or x=4−√32≈2.78 to 2 dec. places