-x^2 + 6<13

[WTF?]

\(\displaystyle -x^2 + 6<13\)

SOOO, I get the solution as 1/2 +- 3 (?3)i

So it has a solution...but it may not be the obvious one...so which is the solution and why?

Thank you!

Also, |x-7|+13 >0

I get it as |x-7|<13 or |x-7|>-13

Therefore the solution (-infinity, 20] and [-6, infinity)

So the solution... :?

 

[mmm4444bot]

Hmmm. Are you sure that you typed both of these exercises correctly?

-x^2 + 6 < 13

-x^2 < 7

x^2 > -7

Any squared quantity is greater than a negative number, so x can be anything real.

|x - 7| + 13 > 0

|x - 7| > -13

The absolute value of any expression is greater than a negative number, so x can be anything real.

Agree? Disagree?

 

[WTF?]

Re:

Hmmm. Are you sure that you typed both of these exercises correctly?

-x^2 + 6 < 13

-x^2 < 7

x^2 > -7

OH, apologies It's actually -x^2 +x+6 <13

|x - 7| + 13 > 0

|x - 7| > -13

The absolute value of any expression is greater than a negative number, so x can be anything real.

Agree? Disagree?

Agree.

Also, pardon me for asking, but why exactly is there no solution to x^2 -3<-7

:?

 

[mmm4444bot]

… why exactly is there no solution to x^2 -3 < -7

Add 3 to both sides of this inequality.

Then, go back to my original response, and re-read my comment about the value of any quantity squared, and see if the reason that you're looking for slaps you in the face.

On your first exercise, I now see why you typed 1/2 +- 3 (?3)i.

Those second terms are also divided by 2, so the zeros of that polynomial are actually:

1/2 + (3/2) * ?3 i and 1/2 - (3/2) * ?3 i

Do you know? When the zeros of a quadratic polynomial are complex numbers with an imaginary part, then the graph has no x-intercepts.

In other words, if a parabola lies entirely above the x-axis or entirely below the x-axis, then there is no value of x that will cause the quadratic polynomial to evaluate to zero. That's why you get imaginary numbers. There are no Real x-intercepts.

By looking at the polynomial x^2 - x + 7, we see that the graph is a parabola that opens upward, right? (When the leading coefficient is positive, the parabola opens upward; when the leading coefficient is negative, the parabola opens downward.)

Therefore, this parabola lies entirely above the x-axis. Since all of the y-coordinates are positive above the x-axis, the value of this polynomial will always be positive.

I'll say it another way.

x^2 - x + 7 = y

Since the parabola is entirely above the x-axis, we know that y is greater than zero for all x. This tells us that the value of the polynomial is greater than zero for all x.

x^2 - x + 7 > 0

The exercise wants to know what values of x make this inequality true. So, the solution is x = all Real numbers.

You can also verify that the lowest part of the parabola is above the x-axis by using the common formula to find the vertex coordinates (i.e., the lowest point on a parabola that opens upward).

x-coordinate of vertex = -b/(2a)

This is 1/2.

Now find the corresponding y-coordinate when x is 1/2.

y = x^2 - x + 7

y = (1/2)^2 - 1/2 + 7

y = 7 + 1/4 - 1/2

This is clearly positive. So, the lowest part of the parabola is definitely above the x-axis. So, the value of the polynomial will always be greater than zero, no matter what value x takes on.

(As you can see, we can approach the first exercise from different directions using all of the facts that we know about quadratics. If this was too much info, and you still don't understand why the answer is all Real numbers, then let me know.)

 

[WTF?]

Wouldn't it be -x^2 +x -7? :idea:

 

[mmm4444bot]

Wouldn't it be -x^2 +x -7 ?

I like working with a positive leading coefficient, so I multiplied both sides by -1. That's why the inequality symbol is reversed. (Did you notice?)