# Homework Help: Find coordinates of A and B

1. Jun 14, 2014

1. The problem statement, all variables and given/known data
The points O,A and B are vertices of an equilateral triangle. Find a and b
O=(0,0)
A=(a,11)
B=(b,37)

2. Relevant equations
$c^2=a^2+b^2$
3. The attempt at a solution
Let AB =c
Then $c=\sqrt{(a-b)^2+(11-37)^2}=\sqrt{(a-b)^2+676}$
Since it is an equilateral triangle, OA and OB is also =c
So $c=\sqrt{11^2+a^2}$
$c=\sqrt{37^2+b^2}$
$\sqrt{(a-b)^2+676}=\sqrt{11^2+a^2}=\sqrt{37^2+b^2}$
I have got to know that $a^2+b^2=1248$, but what next?
I need to find at least one variable to find the other.

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2. Jun 14, 2014

### ehild

You have two independent equations:
$\sqrt{(a-b)^2+676}=\sqrt{11^2+a^2}$ and
$\sqrt{(a-b)^2+676}=\sqrt{37^2+b^2}$, for example.

Square the equations, and solve for a, b.

ehild

3. Jun 14, 2014

### Jilang

You've used two equations for the length of c. There is a third one, can you see it?

4. Jun 14, 2014

### ehild

It is not independent of the other two.

ehild

5. Jun 14, 2014

I have a question.
Why is squaring both sides the same as multiplying both sides by something?
For example,we have an equation such that $a=b$
If we multiply both sides, the equation remain unchanged, $2a=2b$ --> $a=2b/2$--> $a=b$
Again we have an equation such that $a=2b$
If we square both sides, --> $a^2=(2b)^2$ --> $a \times a=2b \times 2b=4 \times b \times b$ --> $a=(4 \times b \times b)/a$

If I do it another way,it remains unchanged: $a=2b$--> $a^2=(2b)^2$ --> $a=\sqrt{(2b)^2}$--> $a=2b$

6. Jun 14, 2014

### ehild

Squaring both sides of an equation is not the same as multiplying both sides with something. The equation stays true, but squaring might introduce false roots. You have to check at the end.
How would you isolate the unknowns without squaring?
If you have the equation a=b and you square both sides, a2=b2, you can isolate one variable by taking the square roots. You get the solution a=±b. a=-b is false.

ehild

7. Jun 14, 2014

Ok, nevermind

I have got this result after rearranging,squaring etc.
$b^2-2ab+676=11^2$
$a^2-2ab+676=37^2$

Before solving the simultaneous equation,I strongly believe this should become
$b^2=11^2$
$a^2=37^2$
but I don't know how. It's not possible to just cancel out the terms since they do not have opposite signs.

8. Jun 14, 2014

### ehild

Check. Plug in the values and see what you get.

ehild

9. Jun 14, 2014

It's wrong -_-
I have never solved equations of this kind. I don't know what to do when I have 3 variables($-2ab,b^2 and a^2$) . Any hints?

10. Jun 14, 2014

### ehild

You have two unknowns, a and b. Isolate b from the second one and substitute the expression for b into the first one. Yo get an equation for a alone.

ehild

11. Jun 14, 2014

$b=\frac{a^2-693}{2a}$
$a^2-2a. \left(\frac{a^2-693}{2a} \right)+676=37^2$
Solving for a, we get $a \in R$

12. Jun 14, 2014

### ehild

You substituted back into the same equation.

$b=\frac{a^2-693}{2a}$
Substitute into
$b^2-2ab+676=11^2$

ehild

13. Jun 14, 2014

Ah,
a=36.37
b=8.66

I solved this using my computer, I should do it myself too. Let me try:
Isolating b
$a^2-2ab=693$
$b=\frac{693-a^2}{-2ab}$
$=\frac{a^2-693}{2a}$

Plug it into the first equation
$b^2-2ab+676=121$
$\left(\frac{a^2-693}{2a} \right)^2-2a\left(\frac{a^2-693}{2a} \right)=-555$
After doing many steps, I got to this:$$-3a^4+3606a^2+480249=0$$

I am not sure what I should do after this step , some kind of 4th power equation?

14. Jun 14, 2014

### SteamKing

Staff Emeritus
Look carefully at the equation you wind up with. It's a quadratic in a^2.

15. Jun 14, 2014

Why should it be a quadratic equation? We have a $-3a^4$ there.

16. Jun 14, 2014

### Ray Vickson

You really don't see what you get when you put $a^2 = x$?

17. Jun 14, 2014

oh, I never thought of it :shy:
This is the first time I have solved such a scary equation-Feeling happy
Thanks guys

btw, can't we solve it using elimination method?

Last edited: Jun 14, 2014
18. Jun 14, 2014

### Ray Vickson

Standard trick. Whenever you have an equation in even powers $a^{\pm 2k}$ you can put $a^2 = x$ and get an equation involving $x^{\pm k}$.

As for 'elimination': yes, you can use the equality $a^2 + 11^2 = b^2 + 37^2$ to write $b$ in terms of $a$. Since your drawing indicates $b \geq 0$ there will not (at this stage) be a problem of choosing the correct sign: $b = \sqrt{a^2 - 1248}$. Put that into the equation $a^2 + 11^2 = (a-b)^2 + 676$, so you have eliminated $b$. However, the result is not a very nice equation for $a$.

19. Jun 14, 2014

No. I mean can we use the elimination method for solving these two equations?(Simultaneous equation)
$(a-b)^2+676}=11^2+a^2$
$(a-b)^2+676=37^2+b^2$

20. Jun 14, 2014

### Saitama

There is a much shorter solution. Use complex numbers. :)

Do you see that the following equation holds?

$$b+37i=(a+11i)e^{i\pi/3}$$

Just compare the imaginary and real parts.

21. Jun 14, 2014

### Ray Vickson

(Error corrected above---you had an unmatched '}').

Which message are you responding to? Please use the 'quote' button, so such confusion is eliminated.

My previous post did, indeed, show how to use the 'elimination' method on those very two equations. However, it first used the fact that the two right-hand-sides are equal---because the left-hand-sides are---to get an equation that makes the elimination much, much easier.

22. Jun 14, 2014

### ehild

It is very clever, Pranav!!

ehild

23. Jun 14, 2014

### Saitama

Thank you, sir!

24. Jun 15, 2014

### thelema418

I think about "squaring both sides" in terms of multiplication with substitution. In your example you have $a = 2b$. If we multiply the equation by $a$, and this yields $a \cdot a = a (2b)$. We can substitute the $a$ on the right-hand side of the equation with $a = 2b$. Now, $a\cdot a = (2b)\cdot (2b)$, or simply $a^2 = (2b)^2$.

Note that where you wrote , you can substitute $a = 2b$. Then $a = (4 \times b \times b) / 2b = 2b$.

I hope that helps.

25. Jun 15, 2014

If we have two equations $x+y=1$ and $x-y=3$
I am having trouble here because the first two terms are different and the $b^2$,$a^2$ and $-2ab$ seems to act like three different variables. Hope you understand what I am trying to say