Yes you ar on the right track but to get from line 1 to line 2 you were multiplying throughout by a^2b^2. So instead of =1 you should have =a^2b^2
ax^2+bx+c=0 has repeated roots if b^2-4ac=0 so apply this to your corrected equation.
Yes you ar on the right track but to get from line 1 to line 2 you were multiplying throughout by a^2b^2. So instead of =1 you should have =a^2b^2
ax^2+bx+c=0 has repeated roots if b^2-4ac=0 so apply this to your corrected equation.
OK, here goes:
a = (b^{2} - m^{2}a^{2})
b = 2mca^{2}
c = (a^{2}b^{2} - c^{2})
For repeated roots, b^{2} = 4ac
4m^{2}c^{2}a^{2} = 4.(b^{2} - m^{2}a^{2})(a^{2}b^{2} - c^{2})
If I try to simplify and solve this equation, I don't get the result b^{2} - 4ac = 0
If I simply ignore the constants a,b & c, I get the equation x^{2}+2x+1 = 0, for which I get the roots (x+1) & (x+1)
Is this correct?
So,
$\displaystyle x^2(b^2 - m^2a^2) + x(-2mca^2) + (-a^2c^2- a^2b^2) = 0$
a = (b^{2} - m^{2}a^{2})
b = -2mca^{2}
c = (- a^{2}c^{2} - a^{2}b^{2})
b^{2} = 4ac
4m^{2}c^{2}a^{4} = 4.(b^{2} - m^{2}a^{2}).(-a^{2}c^{2} - a^{2}b^{2})
4m^{2}c^{2}a^{4} = 4[ -a^{2}b^{2}c^{2} + m^{2}a^{4}c^{2 } - a^{2}b^{4} + m^{2}a^{4}b^{2}]
The two sides are not equal.
We get 4m^{2}c^{2}a^{4} on both sides, but what to do with the other terms?
That 1 in your second line should be a^2b^2 (You were in the process of multiplying both sides by a^2b^2
So line should read (b^2-a^2m^2)x^2-2a^2mcx-a^2(b^2+c^2)=0
Now ax^2+bx+c=0 has repeated roots if b^2-4ac=0
So in our equation want (2a^2mc)^2+4(b^2-a^2m^2)a^2(b^2+c^2)=0
That 1 in your second line should be $\displaystyle a^2b^2$ (You were in the process of multiplying both sides by $\displaystyle a^2b^2$
So line should read $\displaystyle (b^2-a^2m^2)x^2-2a^2mcx-a^2(b^2+c^2)=0$
Now $\displaystyle ax^2+bx+c=0$ has repeated roots if $\displaystyle b^2-4ac=0$
So in our equation want $\displaystyle (2a^2mc)^2+4(b^2-a^2m^2)a^2(b^2+c^2)=0$
I have carried out the correction of multiplying RHS also by $\displaystyle a^2b^2$ and started with the corrected equation,
$\displaystyle x^2(b^2 - m^2a^2) + x(-2mca^2) + (-a^2c^2- a^2b^2) = 0$
which can be rewritten as
$\displaystyle x^2(b^2 - m^2a^2) + x(-2mca^2) -a^2(c^2 + b^2) = 0$
I have carried out the multiplication as you have said. Incidentally, the equation should be
$\displaystyle (2a^2mc)^2-4(b^2-a^2m^2)a^2(b^2+c^2)=0$
and not
$\displaystyle (2a^2mc)^2+4(b^2-a^2m^2)a^2(b^2+c^2)=0$
because the term "C" is negative.
which I have correctly done. How does all this relate to the original equation $\displaystyle p^2 = a^2cos^2 A - b^2 sin ^2 A$?
Edit: Whether the term is negative or positive, it does not matter since there are uncancelled terms on the RHS.
This is correct. Now multiply out brackets. There will be an a^2 which can be cancelled out of every term (since right hand side=0)
You should also see among the terms a^2m^2c^2-a^2m^2c^2, so these go.
Now you should see that there is b^2 which can be cancelled.
Should now have b^2+c^2-a^2m^2=0
Now replace c and m by what we let them stand for at the beginning.