Quadratics, Inequalities and Square Forms

**Yppolitia** · November 2nd 2006, 10:08 AM

Just a few questions I need help on so I can do the rest of my homework.

I have a question which I dont understand very well, the question is 'Solve the following inequalities by sketching by hand the curves of the functions involved' I dont understand exactly what to do with say an equation like p^2-5p+4<0 and 3x^2+5x-2<0

Square Form
How would I do:
2x^2+4x+6

-x^2-2x+5

-x^2+3x+10

**topsquark** · November 2nd 2006, 11:40 AM

Originally Posted by Yppolitia

Just a few questions I need help on so I can do the rest of my homework.

I have a question which I dont understand very well, the question is 'Solve the following inequalities by sketching by hand the curves of the functions involved' I dont understand exactly what to do with say an equation like p^2-5p+4<0 and 3x^2+5x-2<0

Square Form
How would I do:
2x^2+4x+6

-x^2-2x+5

-x^2+3x+10

Well, for starters you haven't given us an inequality to solve...

Let's assume we have $2x^2 + 4x - 6 > 0$ .

The first way to do this is to sketch the graph of $y = 2x^2 + 4x - 6$ and then see where y > 0. By looking at the graph below (hopefully it will load) you can see what the solution set will be: $(-\infty, -3) \cup (1, \infty)$ .

The other way:
We wish to find the critical points of the function $2x^2 + 4x - 6$ . These are the points where:
1) The function is 0.
2) The denominator is 0.
3) The function under the radical is 0.

For this function, the only critical points are where the function is 0. So solve:
$2x^2 + 4x - 6 = 0$ Divide both sides by 2:

$x^2 + 2x - 3 = 0$

$(x + 3)(x - 1) = 0$

So x = -3 and x = 1 are the critical points.

Now we want to break the real line into intervals and test the inequality on each interval:
$(-\infty, -3)$ : $2x^2 + 4x - 6 > 0$ (Check!)
$(-3, 1)$ : $2x^2 + 4x - 6 < 0$ (Nope!)
$(1, \infty)$ : $2x^2 + 4x - 6 > 0$ (Check!)

So we see the solution set is: $(-\infty, -3) \cup (1, \infty)$ as we saw from the graph.

-Dan

**Yppolitia** · November 2nd 2006, 11:53 AM

So for the answer of the quesiotn, please write i completed square form '2x^2+4x+6' what do I put?

**topsquark** · November 2nd 2006, 12:00 PM

Originally Posted by Yppolitia

So for the answer of the quesiotn, please write i completed square form '2x^2+4x+6' what do I put?

Alright, perhaps I didn't understand what you meant. I had thought your question was about how to solve quadratic inequalities. Can you give me an example of what you mean by "write i completed square form '2x^2+4x+6' "

The only other thing I can think of is to rewrite $2x^2 + 4x + 6$ as:
$2x^2 + 4x + 6$

$2(x^2 + 2x + 3)$

$2((x^2 + 2x) + 3)$

$2((x^2 + 2x + 1 - 1) + 3)$

$2((x^2 + 2x + 1) - 1 + 3)$

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

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

This method is called "completing the square." Is this what you mean?

-Dan

**Yppolitia** · November 2nd 2006, 12:17 PM

Yes, thats right

So how would something like -x^2+3x+10 be done because Im not sure how to do an equation with a -x^2 instead of a more common x^2.

**topsquark** · November 2nd 2006, 04:44 PM

Originally Posted by Yppolitia

Yes, thats right

So how would something like -x^2+3x+10 be done because Im not sure how to do an equation with a -x^2 instead of a more common x^2.

What I always do is factor the coefficient of the $x^2$ term:

$-x^2 + 3x + 10$

$-(x^2 - 3x - 10)$

Then isolate the quadratic and linear x terms:

$-((x^2 - 3x) - 10)$

Then add and subtract the constant that completes the square in the parenthesis. Since the coefficient of the $x^2$ term is always 1, take the linear coefficient (-3 in this case), divide by 2 then square it:

$-\left ( \left ( x^2 - 3x + \frac{9}{4} - \frac{9}{4} \right ) - 10 \right )$

$-\left ( \left ( x^2 - 3x + \frac{9}{4} \right ) - \frac{9}{4} - \frac{40}{4} \right )$

Now simplify:
$-\left ( \left ( x - \frac{3}{2} \right )^2 - \frac{49}{4} \right )$

$- \left ( x - \frac{3}{2} \right )^2 + \frac{49}{4}$

-Dan

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