Double discrimiant and ln

[__SB]

Hello all. I have two questions I was hoping you could helpme with.
Q1: (not really afull-on question)
ln ( y – 5) = ln (50/49) – 6
if you e both sides
y – 5 = e ^ (ln (50/49) – 6)
I know the ln and e cancel but was wondering how this gave
Y – 5 = 50/49 * e^-6
Is there a power rule or something going on here?

Q2 : Show that, forall values of the constant k , the equation
tan (x + 60) * tan (x– 60) = K^2 has two roots in the interval 0 < x <180
I first re-wrote this as : (3k^2)tan^2 (x) + tan^2 (x) – 3 –k^2 = 0 (This is correct- looked at ms)
So my thinking was if I used the discriminant I could answerthe question
b^2 – 4ac = 0^2 – (4 * (3k^2 + 1) * -(3 + k^2 ) = 12k^4 +40k^2 + 12
As we have K^4 and K^2 and +12, all these are position forany value of k, hence b^2 – 4ac > 0 thus 2 real roots.
However it apparently wasn’t enough to use the discriminateonce, you had to double discriminant for some reason. Anybody know why?
Thanks in advance.

 

[Deleted member 4993]

Hello all. I have two questions I was hoping you could helpme with.
Q1: (not really afull-on question)
ln ( y – 5) = ln (50/49) – 6
if you e both sides
y – 5 = e ^ (ln (50/49) – 6)
I know the ln and e cancel but was wondering how this gave
Y – 5 = 50/49 * e^-6
Is there a power rule or something going on here? ... yes → a^(m-n) = a^m * a^(-n)

Q2 : Show that, forall values of the constant k , the equation
tan (x + 60) * tan (x– 60) = K^2 has two roots in the interval 0 < x <180
I first re-wrote this as : (3k^2)tan^2 (x) + tan^2 (x) – 3 –k^2 = 0 (This is correct- looked at ms)
So my thinking was if I used the discriminant I could answerthe question
b^2 – 4ac = 0^2 – (4 * (3k^2 + 1) * -(3 + k^2 ) = 12k^4 +40k^2 + 12
As we have K^4 and K^2 and +12, all these are position forany value of k, hence b^2 – 4ac > 0 thus 2 real roots.
However it apparently wasn’t enough to use the discriminateonce, you had to double discriminant for some reason. Anybody know why?
Thanks in advance.

I do not understand the term "had to double discriminant"

 

[__SB]

I do not understand the term "had to double discriminant"

You had to use the discriminant once.
Then use the discriminant on the solution from your first discriminant.
Sorry for the confusion, and thank you for the first bit.

 

[Ishuda]

Hello all. I have two questions I was hoping you could helpme with.
Q1: (not really afull-on question)
ln ( y – 5) = ln (50/49) – 6
if you e both sides
y – 5 = e ^ (ln (50/49) – 6)
I know the ln and e cancel but was wondering how this gave
Y – 5 = 50/49 * e^-6
Is there a power rule or something going on here?

Q2 : Show that, forall values of the constant k , the equation
tan (x + 60) * tan (x– 60) = K^2 has two roots in the interval 0 < x <180
I first re-wrote this as : (3k^2)tan^2 (x) + tan^2 (x) – 3 –k^2 = 0 (This is correct- looked at ms)
So my thinking was if I used the discriminant I could answerthe question
b^2 – 4ac = 0^2 – (4 * (3k^2 + 1) * -(3 + k^2 ) = 12k^4 +40k^2 + 12
As we have K^4 and K^2 and +12, all these are position forany value of k, hence b^2 – 4ac > 0 thus 2 real roots.
However it apparently wasn’t enough to use the discriminateonce, you had to double discriminant for some reason. Anybody know why?
Thanks in advance.

Let's do Q2 first: Collecting terms we have
(3k²+1) tan²(x) - (3+k²) = 0
or, as you would have derived with the quadratic formula had you continued
tan(x) = $\displaystyle \pm\, \sqrt{\frac{3+k^2}{1+3k^2}}\, =\, \pm\alpha$
Now the question becomes how many times can this happen in the given Domain [range of values for x]. What if the Domain had been [0, 360$\displaystyle ^\circ$]?

For Q1, yes there is a power rule:
a^b+c = a^b * a^c