Isn't this the same question you've posted earlier? I've explained example 5 in the comments in the previous post

But I need to explain it using completed the square method.

Which was what I did...

I don’t really understand what you’re doing? I suck at A math

The square of any real number is always 0 or positive. It is never negative.

For example,

5² = 5 x 5 = 25
0² = 0 x 0 = 0
(-3)² = -3 x -3 = 9

So, for any real number x,

x² is always greater or equal to 0.

This is represented by :

x² ≥ 0 (≥ is the 'greater than or equal to' sign)

If x² ≥ 0 , then twice of it also ≥ 0

Which gives us : 2x² ≥ 0

For example,
if x = 7, then 2x² = 2(7²) = 2 x 49 = 98 , which is bigger than 0.

If x = 0, then 2x² = 2(0²) = 2 x 0 = 0

So 2x² ≥ 0.

Now,

If 2x² ≥ 0,

Subtract 4 from both sides of the inequality,

then 2x² - 4 ≥ 0 - 4
2x² - 4 ≥ -4

2x² - 4 ≥ -4, means it is always bigger or equal to -4. So it can never be less than -4.


For example,

If x = -4, then 2x² - 4 = 2(-4)² - 4
= 2 x 16 - 4
= 28, which is bigger than -4.

If x = 0, then 2x² - 4 = 2(0)² - 4
= 2 x 0 - 4
= 0 - 4
= -4
So 2x² - 4 can never be less than -4.

You can try this with different values of x and you will see that it is never less than 4.


Alternatively, go to desmos.com and plot the graph of y = 2x² - 4 to see how this works.


2x² - 4 is already in the completed square form (x - r)² + k. In this case, r =0 and k = -4

Hello Nic.

The expression 2x^2 -4 is already a complete square.

That is why J didn't didn't do anything to the expression other than to show you why it is never less than -4.

Yuan.