Thanks Dr Peterson, I’m on an apprenticeship in engineering and maths is involved a little, but the teacher we’ve been given is new, and not fluent in English and I think he’s finding it difficult to explain mate. So hopefully you can explain better for me to understand ?. On the attached photo where do I go from here, do I rearrange it somehow to get X and Y or do you recommend something else?
Stuck on this question
- Thread starter JackJ1508
- Start date
Your next step is to expand the squared terms in both equations. Start with the numbers. You can use your calculator.
[MATH](64 + 25)^2 = 89^2 = \text {WHAT numerically?}[/MATH]
[MATH](60.5 + 25)^2 = 85.5^2 = \text {WHAT numerically?}[/MATH]
Now expand the algebraic terms.
[MATH](x_2 - 8)^2 + (y_2 - 12)^2 = \text {WHAT algebraically?}[/MATH]
[MATH](x_2 - 152)^2 + (y_2 - 12)^2 = \text {WHAT algebraically?}[/MATH]
[MATH](64 + 25)^2 = 89^2 = \text {WHAT numerically?}[/MATH]
[MATH](60.5 + 25)^2 = 85.5^2 = \text {WHAT numerically?}[/MATH]
Now expand the algebraic terms.
[MATH](x_2 - 8)^2 + (y_2 - 12)^2 = \text {WHAT algebraically?}[/MATH]
[MATH](x_2 - 152)^2 + (y_2 - 12)^2 = \text {WHAT algebraically?}[/MATH]
Hi Jeff,
would it be something like this:
(X^2)^2+(Y^2)^2+208
and
(X^2)^2+(Y^2)^2+23248
mate?
This is correct.
This is not correct.
Here is a handy little formula that is well worth memorizing
[MATH](\alpha + \beta)^2 = (\alpha + \beta)(\alpha + \beta) =\\ \alpha(\alpha + \beta) + \beta( \alpha + \beta) = \alpha^2 + \alpha \beta + \alpha \beta + \beta^2 \implies \\ (\alpha + \beta)^2 = \alpha^2 + 2 \alpha \beta + \beta^2.[/MATH]Apply that to your equations.
Dr.Peterson
Elite Member
- Joined
- Nov 12, 2017
- Messages
- 16,092
First, since X2, which really ought to be [MATH]x_2[/MATH], looks too much like you mean [MATH]x^2[/MATH] (which is what "x^2" means), and takes longer to write, let's just call the coordinates x and y. So the equations are
[MATH](x-8)^2+(y-12)^2=7921[/MATH]
[MATH](x-152)^2+(y-12)^2=7310.25[/MATH]
I'm guessing you meant to type
[MATH](x_2)^2+(y_2)^2+208[/MATH]
but even that is wrong. You evidently changed [MATH](x-8)^2[/MATH] to [MATH]x^2+8^2[/MATH], which is incorrect.
It appears that you have forgotten how to "expand" (multiply out, eliminate parentheses, or whatever it might be called in your region!) We can either use what some call "FOIL", turning [MATH](x-8)^2[/MATH] to [MATH](x-8)(x-8) = x^2 - 8x - 8x + 64 = x^2-16x+64[/MATH]; or use a formula for the square of a binomial: [MATH](a+b)^2 = a^2 + 2ab + b^2[/MATH]. Does any of that sound familiar?
By the way, one reason we need to see your work is to know what you know and what you never learned or forgot, so we can communicate at the right level for you. If I gave you a "complete" solution, but did so assuming you knew what I mentioned in that last paragraph, it probably wouldn't help. So the more you show us, the better we can tune our help to your needs.
Ok thanks Jeff, so in this case it would be:
(X^2-8)(X^2-8)+(y^2-12)(y^2-12)
(X^2-152)(X^2-152)+(y^2-12)(y^2-12)
Hopefully that’s right ?, where would I go from here mate?
Thanks
Like Dr. Peterson, I think the subscripts are confusing you. And we don't need them.
So if what you meant, forgetting the subscripts, is
[MATH](x - 8)(x - 8) + (y - 12)(y - 12) = 7921 \\ (x - 152)(x - 152) + (y - 12)(y - 12) = 7310.25[/MATH]Then you are on track. Just keep going with the algebra.
Jeff,
Before going any further is this right what I’m doing mate. See attached photo
Attachments
Not quite.
[MATH]64 + 144 = 208 \ne 210.[/MATH]
[MATH]23104 + 144 = 23248 \ne 23148.[/MATH]
Use your calculator; it's why God created them. Other than that, you are humming.
Now the next step is to isolate the unknowns on one side of the equation. So you will have what as a result?
Thanks for this but before I go further could you explain how you got them, just so I know mate.
208 = 208 ≠210
23148 = 23248 ≠ 23148
Sorry if I’m being dumb ?
In your last picture in the second line, you had
+64 and +144, which is correct because 64 = 82 + 144 = 122. Do you see that?
Then you apparently added 64 and 144 together in line three (which was the right idea) but got 210, which ain't right.
In the fifth line, you had
+23104 and +144, which is correct because 23104 = 1522 + 144 = 122. Do you see that?
Then you apparently added 23104 and 144 together in line six (which was the right idea) but got 23148, which ain't right either.
You are working on algebra, but you still have to take care with your arithmetic. Which is why I say use a calculator. Arithmetic by hand is BORING and error prone.
+64 and +144, which is correct because 64 = 82 + 144 = 122. Do you see that?
Then you apparently added 64 and 144 together in line three (which was the right idea) but got 210, which ain't right.
In the fifth line, you had
+23104 and +144, which is correct because 23104 = 1522 + 144 = 122. Do you see that?
Then you apparently added 23104 and 144 together in line six (which was the right idea) but got 23148, which ain't right either.
You are working on algebra, but you still have to take care with your arithmetic. Which is why I say use a calculator. Arithmetic by hand is BORING and error prone.
Oh yeah I see all of this now and changed the adding, sorry about that ?. So where would I go now and what happens tho all the Y and X’s?
Attachments
My next step would be to isolate the unknowns. I am not sure it is necessary in this problem, but I like to get arithmetic out of the way. By isolating the unknowns, I mean to get the known numbers on one side of the equation and simplify them to a single number with the unknowns on the other side of the equation. It often make the arithmetic less messy later on, and messy arithmetic I do NOT like.
I shall show you how to isolate the unknowns for this problem, but it is a basic technique that comes up over and over again.
[MATH]x^2 - 16x + 208 + y^2 - 24y = 7921 \implies x^2 - 16x + \cancel {208} + y^2 - 24y - \cancel {208} = 7921 - 208. [/MATH]
DO YOU SEE WHY? This is a basic idea in algebra.
[MATH]\therefore x^2 - 16x + y^2 - 24y = 7921 - 208 = 7713.[/MATH]
Similarly
[MATH]x^2 - 304x + 23248 + y^2 - 24y = 7310.25 \implies x^2 - 304x + 23248 + y^2 - 24y - 23248 = 7310.25 - 23248.[/MATH]
Exact same idea, do you see?
[MATH]\therefore x^2 - 304x + y^2 - 24y = 7310.25 - 23248 = -15937.75.[/MATH]
Now we have a system of two quadratic equations in two unknowns, which is a not something taught in the first few weeks of algebra. Dr. Peterson, however, "saw" a very simple solution for this very simple example. This is not something that is obvious without experience and is not a general solution to systems of quadratic equations. I am not sure I would have "seen" it and would instead have struggled through the depths of the quadratic formula. He suggested in post #17 subtracting the second equation from the first.
[MATH]x^2 - 16x + y^2 - 24y - (x^2 - 304x + 23248 + y^2 - 24y) = 7713 - (-15937.75).[/MATH]
WHY IS THAT TRUE?
That simplifies to
[MATH]x^2 - x^2 - 16x + 304x + y^2 - y^2 - 24y + 24y = 7713 +15937.75 \implies \\ 288x = 23650.75 \implies WHAT?[/MATH]Let's hope I did not screw up. What an ugly problem.
I shall show you how to isolate the unknowns for this problem, but it is a basic technique that comes up over and over again.
[MATH]x^2 - 16x + 208 + y^2 - 24y = 7921 \implies x^2 - 16x + \cancel {208} + y^2 - 24y - \cancel {208} = 7921 - 208. [/MATH]
DO YOU SEE WHY? This is a basic idea in algebra.
[MATH]\therefore x^2 - 16x + y^2 - 24y = 7921 - 208 = 7713.[/MATH]
Similarly
[MATH]x^2 - 304x + 23248 + y^2 - 24y = 7310.25 \implies x^2 - 304x + 23248 + y^2 - 24y - 23248 = 7310.25 - 23248.[/MATH]
Exact same idea, do you see?
[MATH]\therefore x^2 - 304x + y^2 - 24y = 7310.25 - 23248 = -15937.75.[/MATH]
Now we have a system of two quadratic equations in two unknowns, which is a not something taught in the first few weeks of algebra. Dr. Peterson, however, "saw" a very simple solution for this very simple example. This is not something that is obvious without experience and is not a general solution to systems of quadratic equations. I am not sure I would have "seen" it and would instead have struggled through the depths of the quadratic formula. He suggested in post #17 subtracting the second equation from the first.
[MATH]x^2 - 16x + y^2 - 24y - (x^2 - 304x + 23248 + y^2 - 24y) = 7713 - (-15937.75).[/MATH]
WHY IS THAT TRUE?
That simplifies to
[MATH]x^2 - x^2 - 16x + 304x + y^2 - y^2 - 24y + 24y = 7713 +15937.75 \implies \\ 288x = 23650.75 \implies WHAT?[/MATH]Let's hope I did not screw up. What an ugly problem.
Thanks for this that’s really helpful.
So to find x it would be 23650.75/288=82.121 rounded. Where does the y co ordinate fall into this? Thanks again .
I'd probably not round anything until the VERY, VERY end due to the risk of error build-up.
You have equations in x and y. I'd calculate x2 and put those values for x and x2 back into one of the equations and use the quadratic formula to solve for y.
When I was all done, I'd put all those values back into the equations to make sure I had not screwed up.
I realize this is sort of a curt answer, but my grandson has been put to bed, and my son and I are going to have a drink or two before dinner. There are many helpers here who can help you go the rest of the way.
If all the things I have been saying are very new, I suggest a crash course in basic algebra. Khan Academy perhaps?
You have equations in x and y. I'd calculate x2 and put those values for x and x2 back into one of the equations and use the quadratic formula to solve for y.
When I was all done, I'd put all those values back into the equations to make sure I had not screwed up.
I realize this is sort of a curt answer, but my grandson has been put to bed, and my son and I are going to have a drink or two before dinner. There are many helpers here who can help you go the rest of the way.
If all the things I have been saying are very new, I suggest a crash course in basic algebra. Khan Academy perhaps?
Thanks mate, I’m nearly there now ?. Just need to find what Y is by doing:
(y-12)(y-12)=1816.377803
thanks again