Question

I need these:
One function, f(x), with two real rational solutions.
One function, g(x), with two real irrational solutions.
One function, h(x), with two complex solutions.

Answer 1

@SolomonZelman @issy987 @ranga

Answer 2

Does f(x),g(x),h(x) needs to be polynomia with integer coefficients?

Answer 3

No, they do not

Answer 4

I will do the first one:
f(x) = (x - 2)(x - 3) = x^2 - 5x + 6 has two real solutions

Answer 5

Okay, but why does it have two real solutions?

Answer 6

Solution means setting f(x) to zero and solving for x.
(x - 2)(x - 3) = 0 will be 0 when x =2 or x = 3. 2 and 3 are real solutions.

Answer 7

Okay, that you @ranga for explaining that.

Answer 8

The general form of quadratic equation is ax^2 + bx + c = 0

Do you know what the discriminant is for a quadratic expression?

Answer 9

No, I do not. My teacher hasn't really been much of a help.

Answer 10

\[\Large ax ^{2} + bx + c = 0\]
Let D be Discriminant. Then,
\[\Large D = b ^{2} - 4ac\]
The roots or solutions of a quadratic equation are:
\[\Large \frac{ -b \pm \sqrt{D} }{ 2a }\]

If D is positive, the equation will have two real roots
If D is zero, the equation will have two equal roots
If D is negative, the equation will have two complex roots

Answer 11

Okay, wow, this is a whole lot better learning then how my teacher put it.
Are you a teacher?

Answer 12

No.

Answer 13

Should be :)

Answer 14

The first example I gave you has 2 and 3 as solutions. 2 and 3 are real and they are rational numbers.

For the second function g(x), they want two irrational solutions.
Numbers like:
\[\Large \pi, \sqrt{2}, \sqrt{7}\]
are irrational numbers. They are still real numbers but they are irrational because they cannot be put in the form of a fraction such m/n.

So to have two solutions that are irrational we can multiply:
\[\Large g(x) = (x - \sqrt{2})(x - \sqrt{5}) = x ^{2} - (\sqrt{5} + \sqrt{2})x + \sqrt{10}\]

Answer 15

Okay, this is actually starting to make some sense to me now.

Answer 16

\[x=-3+\sqrt{(3)^2-4(3)}\div2\]

Answer 17

So would that be a complex solution?

Answer 18

For two complex solutions choose a, b and c of the quadratic equation such that the discriminant D is negative. The solution to the quadratic equation given earlier requires taking the square root of D. If D is negative you cannot a real square root of D and so the solutions will be complex.

Answer 19

Yes. Except the divided by 2 should be below the entire thing not just the square root of the discriminant.
And also, they want the function h(x). You possibly chose a = 1, b = 3 and c = 3
so h(x) = ax^2 + 3x + 3 will have two complex roots.

Answer 20

Yes, I didnt know how to put the 2 under the equation.

Answer 21

And the + in your solution should be + or -

Answer 22

Okay, thank you so much for your help :) Learned more from you in the past half hour than I do two hours with my teacher

Answer 23

You are welcome. Glad to be of some assistance.

Answer 24

Oh, two other subsets of info that would be useful under the case the discriminant is positive:
If the discriminant is a perfect square there will be two rational real roots
If the discriminant is not a perfect square there will be two irrational real roots

Answer 25

Okay, so how would I convert something like f(x) = (x - 2)(x - 3) = x^2 - 5x + 6 from standard form to vertex form?

Answer 26

You take x^2 - 5x + 6 and complete the square. Are you familiar with it?

Answer 27

What square are you referring to?

Answer 28

Taking ax^2 + bx + c and expressing it in the form a(x - d)^2 + e is usually referred to as "completing the square"

Answer 29

Okay, so I would plug that into the f(x) equation?

Answer 30

You will take the f(x) expression in the standard form and try to express it in the vertex form by completing the square.
x^2 - 5x + 6 = (x - 5/2)^2 - 25/4 + 6
Do you follow this step?

Answer 31

I believe so

Answer 32

That is all there is to it.
Can you try to put this in vertex form: 3x^2 + 2x - 1 ?

Answer 33

I will try

Answer 34

Factor the 3 out of the entire expression first. It will make it easier to work with.

Answer 35

3x^2 + 2x - 1 = 3(x^2 + 2/3x - 1/3)

Answer 36

Okay, so is the answer what you just wrote or will the answer come from that?

Answer 37

By factoring out the 3 you have made a to be 1 in the quadratic expression.
a = 1, b = 2/3, c = -1/3
Once you have made a to be 1, just divide b by 2 to get the term inside the square
here b/2 = (2/3) / 2 = 1/3
so x^2 + 2/3x - 1/3 = (x - 1/3)^2 - 1/9 - 1/3
Follow?

Answer 38

Okay, written like that I now see what is supposed to be done

Answer 39

so x^2 + 2/3x - 1/3 = (x - 1/3)^2 - 1/9 - 1/3 = (x - 1/3)^2 - 4/9
So the original function 3x^2 + 2x - 1 = 3[ (x - 1/3)^2 - 4/9 ]

The reason you take a quadratic function such as ax^2 + bx + c
and put it in the vertex form: a(x - h)^2 + k is:

If you graph the function f(x) = ax^2 + bx + c you will see it is a parabola. One of the frequently asked questions is what is the coordinates of the vertex of the parabola.

So if you put ax^2 + bx + c in the form a(x - h)^2 + k then the vertex will be (h, k).

Answer 40

Okay, so first step in putting f(x) = (x - 2)(x - 3) = x^2 - 5x + 6 into vertex form would be to put x^2 - 5x + 6 in to complete the square

Answer 41

Yes.
Let us take this problem: Find the vertex of the parabola f(x) = 5x^2 + 4x - 1

The first step is factor the 5 out to make a = 1 in the quadratic expression.
5x^2 + 4x - 1 = 5(x^2 + 4/5x - 1/5)

So b here is 4/5. Divide it by 2 to get 2/5. This 2/5 will go as -2/5 inside the square:

Answer 42

5(x^2 + 4/5x - 1/5) = 5[ (x - 2/5)^2 - 4/25 - 1/5]

Answer 43

Okay, So after the\at, what would you do?

Answer 44

Simplify the expression on the right:
5[ (x - 2/5)^2 - 4/25 - 1/5 ] = 5[ (x - 2/5)^2 - 9/25 ]
Take the factor 5 inside (remember, we want the function to look like a(x-h)^2 + k:
5[ (x - 2/5)^2 - 9/25 ] = 5(x - 2/5)^2 - 9/5
h = 2/5 and k = -9/5
So the vertex is: (2/5, -9/5)

Answer 45

Wow, this is defiantly going to pay off on my quiz tomorrow.

Answer 46

Good luck!
I have to go now.

Answer 47

Well, I am absolutely grateful for your help :)

Answer 48

I just noticed there was a sign mistake (+/-) in my earlier answer which I am correcting here.

Let us take this problem: Find the vertex of the parabola f(x) = 5x^2 + 4x - 1

The first step is factor the 5 out to make a = 1 in the quadratic expression.
5x^2 + 4x - 1 = 5(x^2 + 4/5x - 1/5)
So b here is 4/5. Divide it by 2 to get 2/5. This 2/5 will go as +2/5 inside the square.
If the coefficient of the x term is positive, the square term will be of the form:
(x + m)^2
If the coefficient of the x term is negative, the square term will be of the form:
(x - m)^2
5(x^2 + 4/5x - 1/5) = 5[ (x + 2/5)^2 - 4/25 - 1/5 ] = 5[ (x + 2/5)^2 - 9/25 ]
Take the 5 inside: 5(x + 2/5)^2 - 9/5
Compare it to a(x - h)^2 + k
h = -2/5 and k = -9/5
So the vertex is: (-2/5, -9/5)

Answer 49

Find the vertex of 3x^2 + 2x - 1
3x^2 + 2x - 1 = 3(x^2 + 2/3x - 1/3) = 3[ (x + 1/3)^2 - 1/9 - 1/3 ]
= 3[ (x + 1/3)^2 - 4/9 ]
= 3(x + 1/3)^2 - 4/3
Compare it to a(x - h)^2 + k
h = -1/3 and k = -4/3. Vertex = (-1/3, -4/3)