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
Roy Ø.
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)#

Odlisss
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)#

Mahek ☮
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! :)

Jim G.
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"#

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