Tricky Quadratic formula / trig identities

• binbagsss
In summary, the quadratic formula is a mathematical formula used to solve quadratic equations by plugging in the values of a, b, and c into the equation x = (-b ± √(b^2 - 4ac)) / 2a. Common mistakes when using this formula include forgetting to distribute the negative sign and making calculation errors. If the discriminant is negative, the equation will have complex solutions and the imaginary unit i (√-1) should be used. Trigonometric identities are equations involving trigonometric functions that are useful for simplifying expressions and solving equations.
binbagsss
Member warned about posting without the homework template
Im to solve ##(k+l)^{2}e^{-ila}-(k-l)^{2}e^{ila}=0##, for ## k##,

The solution is ##k=l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)##

FIRST QUESTION

So it's a quadratic in k, should be simple enough, my working so far using the quad. formula is ##k= (4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{4l^{2}(e^{-ila}+e^{ila})^{2}-4l^{2}(e^{-ila}-e^{ila})^{2}})/2(e^{-ila}-e^{ila})##
=##(4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{16l^{2}})/2(e^{-ila}-e^{ila})##
... which doesn't give the correct answer.

SECOND QUESTION.

Even after attaining the correct solution, I'm struggling to show that ##l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)##
LHS: ##cos (al) + isin (al) -1 / cos (al) + isin (al) +1=(cos^{2}(al/2)-sin^{2}(al/2)+isin(al/2)cos(al/2)-1)/(cos^{2}(al/2)-sin^{2}(al/2)+isin(al/2)cos(al/2)+1)##

using## cos 2x= cos^{2}x-sin^{2}x##
and ##sin 2x=2sinxcosx,##

but this doesn't seem to be going anywhere...

Any help on either question really appreciated,

Last edited:
binbagsss said:
Im to solve ##(k^{\color{red}{2}}+l)^{2}e^{-ila}-(k-l)^{2}e^{ila}=0##, for ## k##,

The solution is ##k=l(e^{ial}-1)/(e^{ial}+1)=il tan(al/2)##

FIRST QUESTION

So it's a quadratic in k, should be simple enough, my working so far using the quad. formula is ##k= (4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{4l^{2}(e^{-ila}+e^{ila})^{2}-4l^{2}(e^{-ila}-e^{ila})^{2}})/2(e^{-ila}-e^{ila})##
=##(4l^{2}(e^{-ila}+e^{ila})\pm \sqrt{16l^{2}})/2(e^{-ila}-e^{ila})##
... which doesn't give the correct answer.

SECOND QUESTION.

Even after attaining the correct solution, I'm struggling to show that ##l(e^{ial}-1)/(e^{ial}+1)=il \tan(al/2)##
LHS: ##\cos (al) + i\sin (al) -1 / \cos (al) + i\sin (al) +1=(\cos^{2}(al/2)-\sin^{2}(al/2)+i\sin(al/2)\cos(al/2)-1)/(\cos^{2}(al/2)-\sin^{2}(al/2)+i\sin(al/2)\cos(al/2)+1)##

using## cos 2x= cos^{2}x-sin^{2}x##
and ##sin 2x=2sinxcosx,##

but this doesn't seem to be going anywhere...

Any help on either question really appreciated,
The given solution is consistent with there being a typo in your stated problem statement.

It seems there should be no exponent on the k inside the first parenthesis .

SammyS said:
The given solution is consistent with there being a typo in your stated problem statement.

It seems there should be no exponent on the k inside the first parenthesis .
apologies, yes that was a typo,
edited ta.

binbagsss said:
apologies, yes that was a typo,
edited ta.
Does that help you in solving the problem, or are you still stuck?

SammyS said:
Does that help you in solving the problem, or are you still stuck?
The original question remains.
it was a typo.

For the first part, less tedious and error-prone than using your quadratic solution maybe, you have got there a slightly disguised difference of two squares.

If you can't see that, you can also take a term from right to the left hand side and see how you could then best simplify.

Last edited:
epenguin said:
For the first part,less tedious and error-prone than using your quadratic solution maybe, you have got there a slightly disguised difference of two squares.

If you can't see that, you can also take a term from right to the left hand side and see how you could then best simplify.

I think what you are saying is equivalent to the step I took in simpifying the part inside the square root of the quadratic formula, the last line where I get it simplifies to square rt of 16l^{2} ?

binbagsss said:
The original question remains.
Second Question:
Multiply the numerator & denominator by ## \ e^{-i\, l a/2}\ .##

First Question:

Multiply the original equation by ##\ e^{ila} \ ## .

Then it's pretty quick to get difference of squares else, add ##\displaystyle \ (k-l)^{2}e^{2ila} \ ## to both sides & take square root.

The given solution only uses one of the two possible sign choices.

binbagsss
how many variables have your equation?

1. What is the quadratic formula?

The quadratic formula is a mathematical formula used to solve quadratic equations. It is written as x = (-b ± √(b^2 - 4ac)) / 2a, where a, b, and c are coefficients in the quadratic equation ax^2 + bx + c = 0.

2. How do you use the quadratic formula?

To use the quadratic formula, simply plug in the values of a, b, and c from your quadratic equation into the formula and solve for x. Make sure to follow the correct order of operations and pay attention to the signs when simplifying the equation.

3. What are the common mistakes when using the quadratic formula?

One common mistake is forgetting to distribute the negative sign when simplifying the equation. Another mistake is making a calculation error when solving for the square root. It is important to double check your work and use a calculator if necessary.

4. How do you solve a quadratic equation with complex solutions?

If the discriminant (b^2 - 4ac) is negative, then the quadratic equation will have complex solutions. To solve, follow the same steps as with real solutions, but when taking the square root of the discriminant, use the imaginary unit i (√-1) to represent the complex number.

5. What are trigonometric identities?

Trigonometric identities are equations involving trigonometric functions such as sine, cosine, and tangent that are true for all values of the variables within their domains. These identities are useful for simplifying trigonometric expressions and solving equations involving trigonometric functions.

Replies
1
Views
554
Replies
2
Views
1K
Replies
9
Views
1K
Replies
6
Views
2K
Replies
4
Views
2K
Replies
11
Views
397
Replies
2
Views
1K
Replies
1
Views
660
Replies
12
Views
689
Replies
1
Views
1K