Learning Quadratic Functions: Given y=4x^2-12x+3, my instinct is to write 4x^2-3x-9x+3=(4x-3)(x-1)=0

[AncientShape]

Hello, I am just beginning to learn quadratic functions, and even though I feel decent enough at things like factoring polynomials and radicals, I'm a bit lost at the layout of some these equations. I'm using booklets with a lot of great examples that I can follow, but then the practice questions seem to throw in some curve balls and I get flustered at which direction to go. I'm needing to either factor or use root property for this specifically, but the structure of the equation isn't addressed until they move into explaining quadratic formula, which I'm not being asked to use yet, so I feel like I have no good reference on how to solve some of these questions. An example of this would be: y=4x^2-12x+3

My instinct here is to factor this using the trinomial method: 4x^2−3x−9x+3=0
=x(4x−3)−3(3x−1)=0
=(4x−3)(x−1)=0 add 3 to both sides here and divide by 4
=4x−3+3=0+3
=4x−3=0
x=3/4

add 1 to both sides: x−1=0 and x=1

This doesn't feel like the right path and I'm unsure how to tackle that using root property because the examples I look up seem too different in structure? Can anyone suggest videos that have these types of equations so I can see some breakdowns, or screw my head back on properly because I feel like I'm simply overcomplicating this?

Thank you.

 

[Steven G]

4x^2−3x−9x+3=0
=x(4x−3)−3(3x−1)=0
=(4x−3)(x−1)=0

Where did that last line come from???? After all (4x−3)(x−1)= 4x^2 - 7x + 3 NOT 4x^2 -12x+3

What make you think that you can simplify x(4x−3)−3(3x−1) the way it is? What you did after x(4x−3)−3(3x−1)=0 is beyond me.

So how do you do this problem?
Try factoring 4x^2−12x+3. Maybe (2x-?)(2x-?) or (4x-?)(x-?). If that doesn't work, then you need to use a method called completing the square or using the quadratic formula.

I will find you a video for each of these two topics.

 

[Steven G]

For some reason I am having trouble getting the link for the video. Just go to youtube and do a search for quadratic equations by nothing but math proofs.

 

[stapel]

Hello, I am just beginning to learn quadratic functions, and even though I feel decent enough at things like factoring polynomials and radicals, I'm a bit lost at the layout of some these equations. I'm using booklets with a lot of great examples that I can follow, but then the practice questions seem to throw in some curve balls and I get flustered at which direction to go. I'm needing to either factor or use root property for this specifically, but the structure of the equation isn't addressed until they move into explaining quadratic formula, which I'm not being asked to use yet, so I feel like I have no good reference on how to solve some of these questions. An example of this would be: y=4x^2-12x+3

What were the instructions for this equation? What are you supposed to be doing with it?

My instinct here is to factor this using the trinomial method: 4x^2−3x−9x+3=0

On what basis have you replaced the [imath]y[/imath] with a [imath]0[/imath]? Are you supposed to be finding the intercepts of the original function?

=x(4x−3)−3(3x−1)=0

You're correct so far (in your math).

=(4x−3)(x−1)=0 add 3 to both sides here and divide by 4

Um... what the heck just happened?

=4x−3+3=0+3
=4x−3=0
x=3/4

add 1 to both sides: x−1=0 and x=1

This doesn't feel like the right path and I'm unsure how to tackle that using root property ...

What is "root property"? What is the "that" for which you are supposed to be "using root property"?

By the way, the quadratic expression, [imath]4x^2 - 12x + 3[/imath], does not factor over the integers. Are you maybe supposed to be completeing the square...?

 

[AncientShape]

Haha, okay--some of that is me being way off my rocker and trying way too hard to make things fit because I'm lost. I have so much to learn so any suggestions and pointers are seriously welcome. I am supposed to be finding the roots of quadratic equations by either factoring or using root property. I figure the method depends on the setup of the equation, and I'm trying to factor that equation because it looks to be in the form (ax^2+bx+c=0). I am trying not to use the completing the square or quadratic formula because this specific section is not asking for that, but the upcoming sections will be asking for those methods.


Is it appropriate to say :4x^2−12x+3=(2x−1)(2x−3) ? I need to brush up on this clearly, but sometimes it seems like they are not able to be factored further. Thank for the video suggestion, I will seek it out and watch.

 

[AncientShape]

What is "root property"? What is the "that" for which you are supposed to be "using root property"?

By the way, the quadratic expression, [imath]4x^2 - 12x + 3[/imath], does not factor over the integers. Are you maybe supposed to be completeing the square...?

Radical/square root method

Such as:

x^2=9
√x^2 = ±√9
x= ±3

So basically solving quadratic functions by factoring OR this type of method so that the root of a positive term can have a positive or negative result.

 

[stapel]

...I'm trying to factor that equation because it looks to be in the form (ax^2+bx+c=0).

Not all quadratics (just as not all numbers) are factorable. As mentioned earlier, this particular quadratic does not factor over the integers.

Do the instructions tell you to factorize? Or to find the zeroes? Or what? We can't help you do what you need to do until we know what you're actually supposed to be doing. Please provide that information.

Is it appropriate to say :4x^2−12x+3=(2x−1)(2x−3) ?

When you multiply the factors back together, do you get the original polynomial? No? Then the "factorization" of the unfactorable quadratic is probably incorrect.

Also, what is the "method" that your using trying to use?

I need to brush up on this clearly, but sometimes it seems like they are not able to be factored further. Thank for the video suggestion, I will seek it out and watch.

Do you need lessons on multiplying and factoring quadratics?

 

[JeffM]

First off, let's get the nomenclature straight.

[imath]y = 4x^2 -12x + 3[/imath] is an example of a quadratic function, a rule that relates values of x to a value of y.

There are a number of properties that are true of every quadratic function so recognizing that something is a quadratic function gives you important information about the function. One of the most important of those properties is this one: every quadratic function of a real variable with real coefficients has no, one, or two distinct real zeroes or roots.

If and only if z is a zero (root) of f(x), f(z) = 0.

What is the relevance of the zero of a function? These are also called the roots of a function.

Consider the quadratic equation [imath]4x^2 -12x + 3 = 0[/imath]. Notice that this is not a function at all. It is an equation, a statement that some set of arithmetic operations on one or more specific numbers reults in zero. The specific numbers that make the equation true are the solutions to the equation. So the values of the zeroes of the function [imath]f(x) = 4x^2 -12x + 3[/imath] are exactly the same as the values of the solutions of the equation[imath]4x^2 -12x + 3 = 0[/imath].

So there is a very close relationship between the idea of the solutions to an equation and the zeroes of a function, but it helps to have your vocabulary straight.

Second, there is a very simple rule that will tell you how many real roots a quadratic function will have and so how many real solutions a quadratic equation will have.

Let's write a general formula for a quadratic function [imath]f(x) = ax^2 + bx + c[/imath], where a, b, and c are real numbers and a is not equal zero. Then we have the following rule:

[math]b^2 - 4ac > 0,\text {TWO distinct real roots;}\\ b^2 - 4ac = 0,\text {ONE distinct real root; and}\\ b^2 - 4ac < 0,\text {NO real root.}[/math]
What does this rule tell us about [imath]4x^2 - 12x + 3 = 0[/imath].

Let's calculate [imath]d = (-12)^2 - 4 * 4 * 3 = 144 - 48 = 96 > 0.[/imath]

There are two real solutions. Using the square root of d there is an easy way to find those solutions. But apparently you do not know that method.

Third, we would need to know exactly what the instructions for this exercise are to guide you in how to compute the correct answer in the desired manner. Here is a convoluted but still universal way to find the solution, which may help you understand the logic behind the easy way when you learn the easy way.

[math]4x^2 - 12x + 3 = 0 \implies \\ x^2 - 3x + 0.75 = 0 \implies\\ x^2 - 2(1.5x) = - 0.75 \implies\\ x^2 - 2(1.5x) + 1.5^2 = 1.5^2 - 0.75 \implies\\ (x - 1.5)^2 = 2.25 - 0.75 = 1.5 \implies\\ x - 1.5 = \pm \sqrt{1.5} \implies \\ x = 1.5 \pm \sqrt{1.5} = \dfrac{3 \pm 2\sqrt{1.5}}{2} = \dfrac{3 \pm \sqrt{6}}{2}.[/math]

 

[stapel]

[math]4x^2 - 12x + 3 = 0 \implies \\ x^2 - 3x + 0.75 = 0 \implies\\ x^2 - 2(1.5x) = - 0.75 \implies\\ x^2 - 2(1.5x) + 1.5^2 = 1.5^2 - 0.75 \implies\\ (x - 1.5)^2 = 2.25 - 0.75 = 1.5 \implies\\ x - 1.5 = \pm \sqrt{1.5} \implies \\ x = 1.5 \pm \sqrt{1.5} = \dfrac{3 \pm 2\sqrt{1.5}}{2} = \dfrac{3 \pm \sqrt{6}}{2}.[/math]

Unfortunately, we *still* don't know if that is what the original poster (OP) is supposed to be doing. Until OP replies with the instructions, we'll just be guessing.

 

[Steven G]

Is it appropriate to say :4x^2−12x+3=(2x−1)(2x−3) ? I need to brush up on this clearly, but sometimes it seems like they are not able to be factored further. Thank for the video suggestion, I will seek it out and watch.

You tell me if 4x^2−12x+3=(2x−1)(2x−3)!!
Simply multiply out the right hand side and decide. We don't spoon feed on this site. Post back stating if the two sides are equal.

 

[AncientShape]

I'm not trying to be difficult or withhold what my instructions are-I'm just overcomplicating it, confusing myself in the process and should have done more work before posting here. Math sharks, please do not circle. These are my instructions: "Find the roots of the following quadratic equations by factoring or using root property. Keep answers in fraction and/or radical form, if applicable. Reduce any radicals/fractions to simplest form."

Another masked forum member has swooped into my DM's and helped me with some good information, similar to yours Jeffm, so thanks. I do know the square root method, Jeff, but I was trying too hard to factor it because based on my beginners knowledge-I thought that is what that type of equation called for. Folks, I've been out of school for years. Anyway, this is my little breakdown:

Do you need lessons on multiplying and factoring quadratics?

Haha, I mean...yes?
I know you're probably a good person at heart.

 

[stapel]

Math sharks, please do not circle.

Aw; and I just put on my hat for the circling party! ??

These are my instructions: "Find the roots of the following quadratic equations by factoring or using root property. Keep answers in fraction and/or radical form..."

That's useful information; thank you. I will take this to mean that you are required to solve for the zeroes (roots, solutions) of the quadratic by first completing the square, and then taking square roots.

...I was trying too hard to factor it because based on my beginners knowledge-I thought that is what that type of equation called for.

That *is* what quadratics call for, but *only* by those quadratics that actually factor. Most, as in your case, don't. So they've given you a new tool that allows you to do more stuff.

[imath]\qquad y=4x^2-12x+3[/imath]

[imath]\qquad (-12)^2 - (4)(4)(3) = 96[/imath]

[imath]\qquad x = \dfrac{-(-12) + \sqrt{96\;}}{2(4)}[/imath]

[imath]\qquad x = \dfrac{12 + \sqrt{16\cdot 6\;}}{8}[/imath]

[imath]\qquad x = \dfrac{12 + 4\sqrt{6\;}}{8}[/imath]

[imath]\qquad x = \dfrac{3 + \sqrt{6\;}}{2}[/imath]

This looks like you applied the Quadratic Formula. But the instructions specified that you're to complete the square and then take square roots. (Granted, this process leads *directly* to the Quadratic Formula, but you're not yet "allowed" to use the Formula.)

Haha, I mean...yes?

Sometimes looking at an online lesson (in particular, one that differs in presentation from your textbook and instructor) can make a topic suddenly make sense. There is zero shame in asking for lesson links!

* Solving quadratics by square roots (like [imath]x^2 = 9,\;\textrm{ so }\; x=\plusmn 3[/imath])

* Completing the square to solve quadratic equations

* Solving quadratics by completing the square

By the way, when you take the square root of both sides of an equation, you have to put a [imath]\pm[/imath] ("plus or minus") symbol in front of the radical.

Once you're familiar with the process, please make another attempt. Your first steps will be:

[imath]\qquad 4x^2 - 12x + 3 = 0[/imath]

[imath]\qquad 4(x^2 - 3x) = -3[/imath]

[imath]\qquad \left(\dfrac{-3}{2}\right)^2[/imath]

...and so forth.

I know you're probably a good person at heart.

Eh. My kid would beg to differ...