Completing the Square: x^2+2x+5 is Always Positive

[Mo]

Again , its another case of "dont understand the question" (how the hell am i ever going to get the answer :rofl: )

Anway, on with the question (it seems simple...):

"By completing the square, show that $$x^2+2x+5$$ is positive for all real values of $$x$$ "

I have completed the square, my answer:

$$(x+1)^2+4$$

But i don't know what to do next.(the question is in the section of the "discriminant" (b^2-4ac = or > or < 0) by the way )

I think what I am having a hard time with is "real values of x" .What does hat mean?

I would be gratefull for any help, at all!

Regards,
Mo

 

[mahesh_2961]

hai
since your answer is a sum of a positive number (4) and square of a function (x+1) with x belonging to the set of real numbers (all integers,fractions both positive and negative ... or simply
any number which can be plotted on the real axis) that expression will never be negative.
now since this question appears in the discriminant section... we can find b2-4ac of this... which turn out to be negative ...
no real roots ...
implies that the graph y= x^2+2x+5 (which is a parabola facing upwards) never cuts the x-axis .
ie for no value of x the function y has a negative value
...a lengthy proof for a simple problem isn't it ?

regards
Mahesh

 

[Ameen Khan Gauzel]

After completing the expression to the square, it's not wise to determine the discriminant, because if you were to calculate the discriminant, there is no point in completing the expression to the square.

The completed square form is (x+1)^2 + 4

If you replace (x+1) by (p), you will get p^2 + 4

so, let y = p^2 + 4....[1]

From [1], you can draw a sketch of the shape of the curve. You will see that the curve is a parabola above the X-Axis. Then you just explain that values of y are always positive for all real values of x

 

[Mo]

Thank you both.I understand it fully now :)

Regards,
Happy-Mo