Why Does the Inequality 3x2 + 13 < 12x Have No Real Solution?

[Peter G.]

Hi,

Prove that the inequality 3x² + 13 < 12x has no real solution

Is it because:

3x² - 12x + 13 < 0

And, using the quadratic equation we have to square root a negative number, meaning, the answer will be always greater than 0, not smaller?

Thanks,
Peter G.

 

[Mentallic]

Yes, except you're missing one crucial point in your argument. Consider a parabola that is completely under the x-axis, it too will have non-real roots but it is less than zero for all x. Which property of the quadratic makes it such that you can argue that it's completely above the x-axis?

And if you're interested, a better proof (although yours isn't wrong) would be to complete the square instead.

 

[Peter G.]

Ah! Gotcha I think:

3x² - 12x + 13
3 (x² - 4x + 13/3)
3 (x - 4x + 4 -4 + 13 / 3)
3 ((x-2)² - 4 + 13/3)
3 (x-2)² + 1 = 0

The + 1 proves that the graph was transformed upwards.

Is that what you meant?

And, what's the property of the quadratic you mentioned?

Thanks once again,
Peter G.

 

[Mentallic]

Ah! Gotcha I think:

3x² - 12x + 13
3 (x² - 4x + 13/3)
3 (x - 4x + 4 -4 + 13 / 3)
3 ((x-2)² - 4 + 13/3)
3 (x-2)² + 1 = 0

The + 1 proves that the graph was transformed upwards.

Is that what you meant?

Thanks once again,
Peter G.

Good, you completed the square correctly. Now think about what it's telling us, for the expression $$3(x-2)^2+1$$ we can only have a value of 0 or more for the (x-2)² part because any number squared is equal to or more than zero, and the constant multiplier of 3 means that it is in fact going to be a positive number for all x (equal to 0 at x=2). This, coupled with the +1 means it'll have to be more than zero for all x.
You just need to mention that if the value of a in ax²+bx+c is positive, then the quadratic is a , and if it is less than zero, it's a

 

[Peter G.]

Ok, cool! Thanks!

 

[Peter G.]

Oh, if you don't mind, just one last doubt:
Graph the following showing the vertex and x and y intercepts:

y = x² + 3x + 5

First, I found it has no x intercept
The y intercept I found to be + 5
The vertex, by means of the completing the square: (x + 1.5)² + 2.75, I found (-1.5, 2.75)

But I need another point do be able to plot don't I?

 

[Mark44]

You've found that y = 5 when x = 0, and that the vertex is at (-1.5, 2.75). The axis of symmetry is a vertical line through the vertex, so another point on the graph will be at (-3, 5).

However, since you have the equation, you can also get as many points as you want by picking an x value and calculating the y value that goes with it.

 

[eumyang]

Oh, if you don't mind, just one last doubt:
Graph the following showing the vertex and x and y intercepts:

y = x² + 3x + 5

First, I found it has no x intercept
The y intercept I found to be + 5
The vertex, by means of the completing the square: (x + 1.5)² + 2.75, I found (-1.5, 2.75)

But I need another point do be able to plot don't I?

Yes. A parabola has an axis of symmetry. The vertex is on this axis. So draw a dotted line for the axis (x = -1.5), plot the vertex, (-1.5, 2.75), and plot the y-intercept (0, 5). The y-intercept is 1.5 units to the right of the vertex in the x-direction. Because of the symmetry, you'll have another point, 1.5 units to the left of the vertex in the x-direction, that has the same y-coordinate as the y-intercept. What is that point?


EDIT: Never mind, Mark44 beat me to it.

 

[Peter G.]

Nice, thanks for the alternatives, never thought about using the axis of symmetry and I always forget to resort to the simple skill of plugging the numbers in the equations