Are these really quadratic equations? (2/3)x^2 - (2/3)x = 0 and 6x^2 - 1 = 0

C

chijioke

Full Member
Joined
Jul 27, 2022
Messages
352
I was asked to find the sums and products of the these as quadratic equations?1[math].~~\frac{2}{3}x-\frac{2}{3}x=0[/math] [math]2.~~6x^2-1=0[/math]But I am not seeing this as quadratic equations because I know that quadratic equations are of the form[math]ax^2+bx+c=0[/math] and what are presented are not of that form.IMG_20230319_184531.jpgIMG_20230319_184859.jpg
What do you have to say?
 
In your typesetting you missed the square from # 6. As shown in the photo, they are both quadratic equations. Note that in [imath]ax^2+bx+c=0[/imath] only [imath]a \neq 0[/imath], while [imath]b[/imath], or [imath]c[/imath] or both can be zeros.
 
chijioke said:
I was asked to find the sums and products of the these as quadratic equations?1[math].~~\frac{2}{3}x-\frac{2}{3}x=0[/math] [math]2.~~6x^2-1=0[/math]But I am not seeing this as quadratic equations because I know that quadratic equations are of the form[math]ax^2+bx+c=0[/math] and what are presented are not of that form.View attachment 35279View attachment 35280
What do you have to say?
Click to expand...
What you have posted for the first equation differs (significantly) from your picture!

While you posted:- [math].~~\frac{2}{3}x-\frac{2}{3}x=0[/math]Your picture shows:- [math]6  \frac{2}{3}x^2-\frac{2}{3}x=0[/math]
But:- [math]\frac{2}{3}x^2-\frac{2}{3}x=0   \Rightarrow  \frac{2}{3}x^2-\frac{2}{3}x +0=0[/math]and
[math]6x^2-1=0   \Rightarrow  6x^2+0x-1=0[/math]
Does that resolve your "difficulty"?

See also @blamocur's comment(s) above. (Posted while I was 'composing' this response. ?)
 
Last edited:
The Highlander said:
What you have posted for the first equation differs (significantly) from your picture!

While you posted:- [math].~~\frac{2}{3}x-\frac{2}{3}x=0[/math]Your picture shows:- [math]6  \frac{2}{3}x^2-\frac{2}{3}x=0[/math]
But:- [math]\frac{2}{3}x^2-\frac{2}{3}x=0   \Rightarrow  \frac{2}{3}x^2-\frac{2}{3}x +0=0[/math]and
[math]6x^2-1=0   \Rightarrow  6x^2+0x-1=0[/math]
Does that resolve your "difficulty"?

See also @blamocur's comment(s) above. (Posted while I was 'composing' this response. ?)
Click to expand...
The 6 is not included be in the question. It is just a serial number for the question. So the actual thing still remains [math]\frac{2}{3}x^2-\frac{2}{3}x=0[/math] If you notice carefully, you will observe that I used red ink to cover some areas. Those areas contains questions which are not important to our discussion.
 
chijioke said:
quadratic equations are of the form [imath]ax^2+bx+c=0[/imath]
Click to expand...
Hi chijioke. I would reword that statement because Quadratic Equations may appear in other forms:

"Any equation that can be written in the form [imath]ax^2+bx+c=0[/imath] is a quadratic equation".​

In the forms below, the parameters [imath]a,b,c,p,q,h,k[/imath] are Real numbers, and [imath]a[/imath] cannot be zero.

We call [imath]ax^2+bx+c=0[/imath] the Standard Form.

Intercept Form: [imath]~a(x - p)(x - q)=0[/imath], where (p,0) and (q,0) are x-intercepts.

Vertex Form: [imath]~a(x - h)^2+k=0[/imath], where (h,k) is the parabola's vertex.


Later, you may be learning about some other quadratic forms — sometimes called "hidden" forms. Those are not necessarily quadratic equations, but we can solve them after using substitution or simplification to rewrite them in one of the quadratic forms above. Some examples:

\(\displaystyle x^8+7x^4=8\)

\(\displaystyle \bigg(\frac{4}{x}\bigg)\bigg(\frac{x^3}{7}+\frac{2x}{5}\bigg)=\frac{5-x}{35}\)

\(\displaystyle x = \sqrt{x} + 12\)

\(\displaystyle \frac{1}{(3x+1)^2}+5=\frac{6}{3x+1}\)

\(\displaystyle 5^{2x+1}-26(5^x)+5=0\)

Again, you don't need to consider hidden forms now, but be aware that other quadratic forms may be in your future. :)
[imath]\;[/imath]
 
chijioke said:
The 6 is not included be in the question. It is just a serial number for the question. So the actual thing still remains [math]\frac{2}{3}x^2-\frac{2}{3}x=0[/math] If you notice carefully, you will observe that I used red ink to cover some areas. Those areas contains questions which are not important to our discussion.
Click to expand...
Oh dear!

You appear to be completely ignoring the advice you have been given, @chijioke!

I included the "6" simply because it was in your picture. I know only too well that it was the 'question number' and did not seek in any way to include it as part of the equation, as evidenced by the line below where (with no further mention of the "6") I illustrated the fact that the equation given can be restated in the form: \(\displaystyle ax^2+bx+c=0\) (that you failed to recognize for yourself!).

I see that others are now offering further 'assistance' which is their privilege should they choose to do so, of course, but I don't see why anyone should spend their time doing so when you are so clearly refusing to accept the advice already provided!

There are none so blind as those who will not see!
 
The Highlander said:
… the equation given can be restated in the form: \(\displaystyle ax^2+bx+c=0\) (that you failed to recognize for yourself!).

I see that others are now offering further 'assistance' which is their privilege should they choose to do so, of course, but I don't see why anyone should spend their time doing so when you are so clearly refusing to accept the advice already provided!

There are none so blind as those who will not see!
Click to expand...
I think some of those comments are too harsh. :(
[imath]\;[/imath]
 
mmm4444bot said:
I think some of those comments are too harsh. :(
[imath]\;[/imath]
Click to expand...
Maybe, if I was dealing with a primary school pupil but @chijioke has already demonstrated some more advanced capability than that yet chose to criticize my inclusion of the question number that was clearly displayed in their picture as If I didn't know what it was and suggested I wasn't reading their post as "carefully" as I might!
I think I am entitled to be indignant at what was posted at #4 especially when @blamocur had explained the misconception the OP was suffering and I had actually shown (clearly) how to resolve it! ?
 
(2/3)x^2 - (2/3)x=0. I would divide by 2/3 getting x^2 - x =0 or x^2 = x.
Then I would ask myself which number or numbers, if any, when I square them gives me back the same number. This should be easy to think through.
 
Steven G said:
(2/3)x^2 - (2/3)x=0. I would divide by 2/3 getting x^2 - x =0 or x^2 = x.
Then I would ask myself which number or numbers, if any, when I square them gives me back the same number. This should be easy to think through.
Click to expand...
I have no problem solving it. I am only thinking that [math]\frac{2}{3}x^2-\frac{2}{3}x=0[/math] is not a quadratic equation. That is the only argument I have.
 
chijioke said:
I have no problem solving it. I am only thinking that [math]\frac{2}{3}x^2-\frac{2}{3}x=0[/math] is not a quadratic equation. That is the only argument I have.
Click to expand...
Of course it is and, as has already been demonstrated (& explained), can easily be rewritten in the Standard Form of a quadratic (\(\displaystyle ax^2+bx+c=0\)) that you appear to (mistakenly) believe is what defines(?) a quadratic equation! It also produces a 'nice' parabola (as I trust you would expect from a quadratic equation) whose roots can clearly be seen to lie at 0 & 1 (as @Steven G hinted at earlier):-

Parabola.png​
chijioke said:
In other words zero can also be regarded as constant.
Click to expand...
Zero is "constant" isn't it? Or is a constant or, perhaps more appropriately in this case, is a perfectly acceptable value for the coefficient of the \(\displaystyle x^0\) term of a second-order polynomial to take! Such equations being more commonly known as "quadratic" because they contain a 'squared' term (and no higher degree terms or higher degree terms whose coefficients are all zero) and thus must have at least one solution (real or complex); in this case having two (both real) as illustrated above.
 
Last edited:
The Highlander said:
Of course it is and, as has already been demonstrated (& explained), can easily be rewritten in the Standard Form of a quadratic (\(\displaystyle ax^2+bx+c=0\)) that you appear to (mistakenly) believe is what defines(?) a quadratic equation! It also produces a 'nice' parabola (as I trust you would expect from a quadratic equation) whose roots can clearly be seen to lie at 0 & 1 (as @Steven G hinted at earlier):-

Click to expand...
In standard form [math]\frac{2}{3}x^2-\frac{2}{3}x=0[/math] would be written as [math]2x^2-2x-0=0[/math]. Right? If this is correct, then I must confess that I am seeing this kind of quadratic equation for the first time.
 
chijioke said:
In standard form [math]\frac{2}{3}x^2-\frac{2}{3}x=0[/math] would be written as [math]2x^2-2x-0=0[/math]. Right? If this is correct, then I must confess that I am seeing this kind of quadratic equation for the first time.
Click to expand...
No!

It should not be rewritten as: \(\displaystyle 2x^2-2x-0=0\). (Why would you change the coefficients???)

I showed you earlier (in post #2, qv) that it may be rewritten as: \(\displaystyle \frac{2}{3}x^2-\frac{2}{3}x+0=0\)

It doesn't matter whether you add or subtract the zero (because you are adding/subtracting nothing) but changing the coefficients produces a different function! (Maybe try graphing them for yourself in Desmos?)

I also attempted to get you to think of a quadratic as a 2nd-order polynomial so you could think of

\(\displaystyle \frac{2}{3}x^2-\frac{2}{3}x+0\)​

as being the same as

\(\displaystyle \frac{2}{3}x^2-\frac{2}{3}x^1+0x^0\)​

(Note too, that I do not equate the expressions to zero as that is simply a way of determining their roots, ie: where they cross the x-axis if they indeed do so)

Thinking of (all) expressions like this as polynomials where the highest power of \(\displaystyle x\) determines their order/degree, will stand you in good stead when it comes to topics like synthetic division where it is important to recognize and take account of the coefficient of every degree (power) of the variable (usually \(\displaystyle x\)) even when its coefficient is zero!

That catches out many students because if the coefficient is zero then the term simply doesn't "appear" in the expression (normally).

For example:- [math]5x^4-4x^3+2x-1[/math]is really:- [math]5x^4-4x^3+0x^2+2x^1-1x^0[/math]
The x-squared term (\(\displaystyle 0x^2\)) doesn't (usually) "show up" in the expression but you should still consider it as being there, it just has a zero coefficient.

Hope that helps.
 
Correction:-

I should really have written:- [math]5x^4-4x^3+0x^2+2x^1-1x^0[/math]as:-[math]5x^4+ˉ4x^3+0x^2+2x^1+ˉ1x^0[/math]
(because a polynomial should be thought of as a sum of terms where each term's coefficient may be positive or negative or zero.)
 
chijioke said:
In standard form [math]\frac{2}{3}x^2-\frac{2}{3}x=0[/math] would be written as [math]2x^2-2x-0=0[/math]. Right? If this is correct, then I must confess that I am seeing this kind of quadratic equation for the first time.
Click to expand...
You seem to be suffering from perhaps several misapprehensions.

First, the letters a, b, and c in the standard form of a quadratic expression represent numbers. Fractions are a kind of number. There is nothing in that form that limits a, b, or c to integers. You may have been misled by examples that were restricted to integer coefficients.

Second, as several have pointed out, the standard form is not a definition. It is a convenient and therefore frequent way to present a quadratic expression, but it is not the only way to do so. Any quadratic can be expressed in standard form, but it need not be.

Third, the idea of adding zero to an expression as a step in finding a mathematically equivalent but more convenient expression is a quite common technique (as is multiplying an expression or part of an expression by one). Add it to your tool box.

Fourth, be careful to distinguish between equations that describe a function and those that define a solution set.

[math] f(x) = \frac{2}{3} x^2 + \frac{2}{3}x \text { and } g(x) = 2x^2 + 2x \implies \\ f(x) \text { IS NOT GENERALLY EQUAL TO } g(x).\\ \text {BUT}\\ \frac{2}{3}x^2 + \frac{2}{3}x = 0 \implies 2x^2 + 2x = 0. [/math]
 
You must log in or register to reply here.
Top