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

This is what I got up to:
y=2(x4)23
0=2(x4)232
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 x1=4+32 and x2=432

Explanation:

As a2b2=(a+b)(ab), you could write the function as:
y=[2(x4)3][2(x4)+3]

y=0 whenever either of the parentheses is 0, i.e.
2(x4)3]=0, i.e x1=4+32
or 2(x4)+3]=0, i.e. x2=4+32

Jun 12, 2018

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(x4)23
3=2(x4)2
32=(x4)2

Then :

(x4)2322=0
(x432)(x4+32)=0

so x=4+32 or x=432

Jun 12, 2018

x=4±32

x=4+32orx=432

Explanation:

y=2(x4)23

Solving for the x set y=0

0=2(x4)23

2(x4)2=3

(x4)2=32

x4=32

x=4±32

x=4+32orx=432

~Hope this helps! :)

Jun 12, 2018

x=4±32

Explanation:

to solve for x set y=0

2(x4)23=0

add 3 to both sides

2(x4)2=3

divide both sides by 2

(x4)2=32

take the square root of both sides

(x4)2=±32note plus or minus

x4=±32

add 4 to both sides

x=4±32exact values

x=4+325.22 to 2 dec. places

or x=4322.78 to 2 dec. places