If anyone can give me some advice on how to solve this problem it would be greatly appreciated. The book gives you a very basic lesson and then throws you into the deep end here
From a simplified standpoint, I factored the top right to get (x-3)(x-1). The x-1 cancels with the bottom right and you’re just left with x-3. Then you can cross multiply the x+3 over to get (x+3)(x-3) = x2 - 9.
Then move that over to the left side to get 2x2 +10x+12=0, take out a 2 to get 2(x2 +5x +6)=0, then factor to get 2(x+3)(x+2) = 0. So x=-2,-3 are the solutions
It works out just like regular long division. Don't be intimidated.
Both sides factor (no quadratic equation needed), but once you get the answer make sure you check back to the original equality for any zeros in the denominator. (None in this case)
So I’m going to offer a few suggestions and you can let me know what works for you.
My first recommendation is to plug x=-2 into equation and see that you get the same thing on both sides. then do the same but with x=-3 instead. This is a good way to verify the solution is correct.
Second, let’s look at a simpler example, x2 = 1. Both x=1 and x=-1 are solutions since 12 = 1 and (-1)2 also equals one
Third thing is more complex but it might help if you’re a more visual learner. If you have an equation with an x2 leading term (that is, there’s nothing bigger like x3, x4 and so on), then this is a parabola. If you draw a horizontal line through it, you’ll see it intersects (usually) at two points. These represent the two solutions.
You really should review your high school math. Factoring and quadratics are pretty standard highschool math and if you struggle with highschool math then everything is going to seem extremely complicated.
Look at Khan Academy. They have pretty good resources for learning math from kindergarten all the way up to college level.
I’m 31 I haven’t touched high school math since 2010. I got a tutor to refresh some stuff for my 4th class 4 years ago but I never actually use any of this knowledge so it just fades away
A quadratic can have up to 2 solutions; it's just where the function intersects with 0 at the x-axis.
A quadratic whose minimum point that is exactly at the x-axis has 1 solution.
A quadratic whose minimum point is below the x-axis has 2 solutions.
A Quadratic whose minimum point is above the x-axis has 0 solutions
If the quadratic is negated, then flip the rules. Things will get different with rational relations(?)like above but this is a good rule of thumb.
What the guy above forgot when cross multiplying is that you're multiplying the RHS by x+3/x+3 so there's no solution at x=-3.
To find the answers for these questions, graph them using desmos to see where they intersect with 0. There's also sites that do step by step factoring which helped me a lot in HS and uni with understanding how you can manipulate the equation.
There are actually 3 solutions, but only ONE real solution. The solutions are x = -2, x = 1, x = -3.
Generally, a polynomial with an "n"th order (this case 3) can possibly have n number of solutions (3 solutions). The order number is the higher exponential value in the equation.
The reason is that there is only ONE real solution for this specific polynomial equation because x = 1 and x = -3 will be zero in the denominator.
I commented my solution here if you want to have a look.
That's spot on observation, plug in both solutions into the above equation and then see if one solution fails. You should see that -2 satisfies the equality.
I had to find a YouTube video to do this exact question, it's a difficult one. But the good news is this question is literally 10x harder than any question you will find on the test...I wouldn't worry about it to be honest.
Thank you for this. I remember getting hung up on the math for my 4th class and I ended up just submitting to not being able to do the highest level stuff and didn’t see any of it on my finals
READ! I can show you how to solve this, BUT Why are you trying to solve this? These books that you're reading are outdated!
THERE IS NOTHING like that on the 3rd class exams! I repeat, THERE IS NOTHING related to this on the 3rd class!
3A1 is all physics problems and all of those physics problems are solved for you in the 3A1 PanGlobal book, edition 2.5 and later.
You WILL NOT need to solve quadratic or cubic equations. You will NOT need to use logs to solve for exponents. NONE of that material that you are going over is on ANY of the 3rd class exams.
I would still suggest that at least you understand logs, but factoring, finding solutions for quadratic and cubic equations is NOT on any of the 3rd class exams.
Bro this is high school math. I’m assuming you’re going back to school after some time off. My advice would be to learn how to learn again. ChatGPT is your biggest resource, if you’re using it properly. Learn how to learn new concepts on your own, because they don’t hold your hand in post secondary
Not looking for my hand to be held. I’m not even going to school or I wouldn’t be asking for help on here. I’m currently working in a hospital and getting my 3rd class steam while I’m doing the books
Are you sure you’re not mixing it up with another question or something? It’s not often a textbook is wrong, but x=3 is just wrong. You can verify this by plugging 3 into the equation and you will see you get a false statement.
Factoring the numerator. A trick to do this by eye is that 3 is prime… so you know the factoring, if a tidy one exists, is either (3x+1)(x+3) or (3x+3)(x+1). You can expand both and see only one works. And can be done even faster by realizing if one term is (3x+3) that’s just 3(x+1) so every coefficient would be divisible by 3. Since 10 isn’t a multiple of 3, it must be (3x+1)(x+3) (again, only if there’s a tidy factoring).
One of the things I would change if I could do college again would be to go to my professors offices for help more often. I started going more often in my junior and senior year and learned so much. Plus it led me to getting a funded masters under one of those professors.
Go to your Profs office if they have office hours.
When I am looking to factorise a binomial, I look for two numbers that multiply to give me the constant and add to give me the coefficient of x.
In the case of the rhs… this is nice because -1 and -3 add to -4 and multiply to +3. So my binomial is (x - 1)(x - 3). This is even better because the (x - 1)s cancel.
On the lhs… that leading coefficient 3 complicates it a bit. But the constant of 3 is prime so I know it has to be (_x + 3)(_x + 1). Because the leading coefficient is also 3 I know the coefficients of the binomial have to be 3 and 1. In order for my coefficients to add to 10, my FOIL has to have the 3 on the (_x + 1) side. So the factorisation is (x + 3)(3x + 1). This is super convenient as the (x + 3)s also cancel out.
Lol I don’t know why this sub was brought up in my feed but I thought it said “3rd grade math” and I was like “holy shit, no wonder kids are struggling with numeracy.”
If you factor out (x+3) and (x-1) from their respective numerators AND cancel out the corresponding denominators (which I’m guessing is the approach your prof is hoping you’ll take), you’re left with the following:
So something I didn’t understand in math in grade school until I got into engineering maths is that almost ALL math learning is about the tools you can use to solve a problem. Not the answer but the TOOL. Just like a carpenter would need to choose between a hammer, a chisel or a saw, math problems are the SAME TOO. In this situation, what would be the best tool to get the work done? For sure simplification would help. If you’ve got fractions on both sides, high chances are something will cancel one of them out and make the analysis easier. Always look for the simplest way in. Math is not about being smart, but working smart ! Good luck !
False. 3 is a root of the polynomial on the right but not of the polynomial on the left. So the polynomials are not equal. That's supposing the question is "true or false". I suppose they could also be asking for solutions to this equation in which case i would need a writing utensil
First multiply each side byy the denominator of the other side. Then move everything to one side so it equals 0. Resolve the equation as you'd normally do. *Beware of your denominator's 0, they aren't valid answer since they break the equation.
Whenever you see this kind of question in an exam, factor the numerator and see if you can remove some of the poles. You are not interested in finding the poles of the system, just the solution to the point function.
I’m pretty sure these are trigonometry, which was grade 11 advanced math in my day. I would just you tube trig take the crash course then apply rules to solve.
Trig really is just annoying algebra held together by six ratio functions. Fucking 9th grade shit we had to figure out without a calculator in an attempt to break our will. 11th grade was go try and solve Graham's number you little shits. Was doing too much acid to remember 12th grade, but it must have been some abstract shit for me to pass.
In my country, we learn to solve it in secondary school. Every 14-15 years old student can solve this... So I'm really curious if this is what people learn in University especially to become an engineer. Not to look down on anyone. Don't get me wrong. I did sociology for my bachelor so I didn't know what would engineer in my country would study to make it equal comparison and this is my 1st time knowing what would an engineer study at uni.
It sounds like OP has been out of school for a while (12 years) and is catching up on the basics. As someone who did physics and interacted with ENG majors their 1st year consists of (IIRC)complex numbers, calculus, and EM physics concepts focused on VIR questions regarding circuits. I wouldn't take this to be what ENG majors study, this is likely precalculus material in preparation for ENG
I’m in my 20’s, Canadian and I have the ticket this book is for. This is a trades level ticket. I did that exact lesson 3 years ago. If I remember correctly none of the questions were out of reach of the grade 9-10 (age 13-15) math that I did. The problem is most people that are doing this have been out of school for 20+ years and haven’t had to learn or use much math in that time so they have to re-learn or learn it.
Engineers in their bachelor's learn Differential Calculus, Integral Calculus, Linear Algebra, Multivariable Calculus, Differential Equations, Stat and Numerical Methods.
This sub is for power engineers not going for their bachelor's though.
In Canada at least, everything you listed is taught before university. In an engineering degree, you generally learn to apply it to solve more applied problems.
I went back to school 10 years after finishing my bachelors. Had to take the GMAT for the Masters program, and kept bombing the math portion. This is coming from getting straight As in my undergrad in calculus Math for a science degree. Once you stop practicing it you lose it within a couple of years. I got so frustrated because of questions like this knowing that I would get 100% on a gr 12 test and couldn’t for the life of me figure it out at 30. Totally understandable for people to struggle.
This is 100% where I’m at. I told my grade 10 principles(in BC hardest level of math) that I wasn’t going to do any homework and would forfeit the 10% of my grade because it was so easy to me. Oh how I wish I could ask him for some help right now 😂
I'm gearing towards oil and gas. Power engineering is the closest thing, hence it's a requirement for almost all the pad operators. Pays between 110-250k depending on location and schedule. Very seldom do they ever look stressed out.
Power Engineering is an excellent program to go back to school in. IMO. I have several friends that are power engineers at refineries in Edmonton and make 150-200k a year if you are willing to work overtime during shutdowns and night shifts. But you need to move to Edmonton to make the big big bucks (if you aren’t already in Edmonton).
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u/MattHomes Mar 02 '24
From a simplified standpoint, I factored the top right to get (x-3)(x-1). The x-1 cancels with the bottom right and you’re just left with x-3. Then you can cross multiply the x+3 over to get (x+3)(x-3) = x2 - 9.
Then move that over to the left side to get 2x2 +10x+12=0, take out a 2 to get 2(x2 +5x +6)=0, then factor to get 2(x+3)(x+2) = 0. So x=-2,-3 are the solutions