#Help needed!
lean plover
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leaden apex
Notice how you have some similar triangles here.
lean plover
Notice how you have some similar triangles here.
Yes. One above Square and one at the side.
leaden apex
Yes. One above Square and one at the side.
Right.
So, try doing this. Express y in terms of x first.
lean plover
Right.
So, try doing this. Express y in terms of x first.
I have been doing this for quiet a long time and just got lost in variables.
😭😭😭
leaden apex
I have been doing this for quiet a long time and just got lost in variables.
Well, yeah, the equation does become quite complicated... Let's see.
From this we can see:
x/a = a/y
So, y = a^2/x.
Thus, we get an equation:
(x + a)^2 + (a^2/x + a)^2 = l^2
Let's expand it and order the terms.
x^2 + 2ax + a^2 + a^4/x^2 + 2a^3/x + a^2 = l^2
x^4 + 2ax^3 + (2a^2 - l^2)x^2 + 2a^3 x + a^4 = 0
I see that we have a in a lot of places, so let x = at.
a^4 t^4 + 2a^4 t^3 + (2a^2 - l^2)a^2 t^2 + 2a^4 t + a^4 = 0
We can now cancel by a^4, also letting m = l/a for convenience:
t^4 + 2t^3 + (2 - m^2)t^2 + 2t + 1 = 0
Now, do you see anything nice about this equation?
lean plover
Grant me a few minutes.
leaden apex
Of course.
leaden apex
Yeah, nice! It's a palindromic polynomial.
So, dividing by t^2 and grouping gives:
(t^2 + 1/t^2) + 2(t + 1/t) + (2 - m^2) = 0
So, what do we do next?
lean plover
(t + 1/t)^2 - 2?
leaden apex
Right. So, y^2 - 2.
Thus, our equation becomes:
y^2 + 2y - m^2 = 0
And this is now easy.
Just some algebra is left.
By the way, when finding y, note that y > 0 only for t > 0, so you can immediately discard the negative root.
lean plover
Yes, you're right. The value of m should be positive because l and y are positive.
leaden apex
Well, m is obviously positive, since m = l/a.
So, now just solve for y, then for t, then express x = at and substitute m = l/a, and then you can substitute the given values.
lean plover
I cannot think further.....
Umm....wait-
Can we solve by quadratic?
We will get two values of M.
Then we can use the positive value of M?
Am I correct?
leaden apex
We will get two values of M.
You're solving for y. m is just a parameter that contains the given lengths.
And yes, you can now just solve the quadratic equation.
leaden apex
Well, you can take as much time as you want 😄
lean plover
m belongs to (-1,1)
leaden apex
No.
m is always bigger than 1.
That doesn't really matter, though. Just solve for y.
lean plover
Oh, sorry. Wait again? Let me use my brain further. 🙂
lean plover
That doesn't really matter, though. Just solve for y.
Sorry for wasting your time, but I give up now.
I am still trying.
leaden apex
Well, I mean, just use the quadratic formula:
y = -1 ± √(m^2 + 1)
We discard the negative root for reasons above, so:
y = √(m^2 + 1) - 1
Returning to t, we get:
t + 1/t = √(m^2 + 1) - 1
t^2 - (√(m^2 + 1) - 1)t + 1 = 0
And this is another quadratic equation. Here both roots are needed, though. You'll see why later.
lean plover
OHHHHH
lean plover
Well, I mean, just use the quadratic formula:
y = -1 ± √(m^2 + 1)
We discard the...
I literally came up with this....
lean plover
Returning to t, we get:
t + 1/t = √(m^2 + 1) - 1
t^2 - (√(m^2 + 1) - 1)t + 1 = 0...
But I forgot about this, sorry sir.
😭
leaden apex
I'm not that old 😅
No worries!
lean plover
I'm not that old 😅
No worries!
Can I ask you for your age, sir?
leaden apex
I'm 25.
lean plover
Big bro, I am sorry for wasting your time but I am tired of solving the quadratics....
And then put all the value the substitute value.
Can we do it tomorrow?
I guess I should take some rest.
leaden apex
Well, it's your question, you can do it whenever you want.
lean plover
Thanks, big bro. Let me tell you, I have been solving this question for 3 hours from now.
😭
Let me take some rest and we will resume from tomorrow.
Thanks for the assistance that you have given.
leaden apex
You're welcome!
lean plover
Goodnight for now, big bro.
ornate agateBOT
@lean plover