How do you work out the x intercepts of the equation y=2(x-4)^2-3y=2(x4)23 by factorising? Thanks!

This is what I got up to:
y=2(x-4)^2-3y=2(x4)23
0=2(x-4)^2 - sqrt3^20=2(x4)232
0=2(x-4 -sqrt3)(x-4+sqrt3)0=2(x43)(x4+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!!

4 Answers
Jun 12, 2018

The x intecepts are x_1=4+sqrt(3/2)x1=4+32 and x_2=4-sqrt(3/2)x2=432

Explanation:

As a^2-b^2=(a+b)(a-b)a2b2=(a+b)(ab), you could write the function as:
y=[sqrt2(x-4)-sqrt3][sqrt2(x-4)+sqrt3]y=[2(x4)3][2(x4)+3]

y=0y=0 whenever either of the parentheses is 0, i.e.
sqrt2(x-4)-sqrt3]=02(x4)3]=0, i.e x_1=4+sqrt(3/2)x1=4+32
or sqrt2(x-4)+sqrt3]=02(x4)+3]=0, i.e. x_2=4+sqrt(3/2)x2=4+32

Jun 12, 2018

x = 4+-sqrt(3/2)x=4±32

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 - 30=2(x4)23
3 = 2(x-4)^23=2(x4)2
3/2 = (x-4)^232=(x4)2

Then :

(x-4)^2 - sqrt(3/2)^2 = 0(x4)2322=0
(x-4 - sqrt(3/2))*(x-4 + sqrt(3/2)) = 0(x432)(x4+32)=0

so x = 4 + sqrt(3/2)x=4+32 or x =4 - sqrt(3/2)x=432

Jun 12, 2018

color(red)(=> x=4+-sqrt(3/2)x=4±32

color(magenta)(=>x=4+sqrt(3/2) or x=4-sqrt(3/2)x=4+32orx=432

Explanation:

y=2(x-4)^2-3y=2(x4)23

Solving for the x set =>y=0y=0

=>0=2(x-4)^2-30=2(x4)23

=>2(x-4)^2=32(x4)2=3

=>(x-4)^2=3/2(x4)2=32

=>x-4=sqrt(3/2)x4=32

color(red)(=> x=4+-sqrt(3/2)x=4±32

color(magenta)(=>x=4+sqrt(3/2) or x=4-sqrt(3/2)x=4+32orx=432

~Hope this helps! :)

Jun 12, 2018

x=4+-sqrt(3/2)x=4±32

Explanation:

"to solve for x set "y=0to solve for x set y=0

2(x-4)^2-3=02(x4)23=0

"add 3 to both sides"add 3 to both sides

2(x-4)^2=32(x4)2=3

"divide both sides by 2"divide both sides by 2

(x-4)^2=3/2(x4)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"(x4)2=±32note plus or minus

x-4=+-sqrt(3/2)x4=±32

"add 4 to both sides"add 4 to both sides

x=4+-sqrt(3/2)larrcolor(red)"exact values"x=4±32exact values

x=4+sqrt(3/2)~~5.22" to 2 dec. places"x=4+325.22 to 2 dec. places

"or "x=4-sqrt(3/2)~~2.78" to 2 dec. places"or x=4322.78 to 2 dec. places