School Olympiad problem: 25/sqrt[x-1] + 4/sqrt[a-2] = 14 - sqrt[x-1] - sqrt[a-2]

[GilgameshProject]

. The task is to find, for which values of parametr "a" this function has at least 1 solution.

I've encountered this problem and can't solve it despite thinking and trying for several hours. If someone has a clue how to do it, pls help. I have tried to substitute (x-1)^0,5 for another variable and analyze it as a quadratic function with coefficients including paramet "a", but discriminant seems to be really complex and I don't know what to do with it. Thanks in advance :<

 

[mmm4444bot]

Hello. Is this exercise from an active competition?
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[GilgameshProject]

Hello. Is this exercise from an active competition?
[imath]\;[/imath]

No, it is from 2021

 

[Steven G]

Replace sqrt(x-1) with u and sqrt (a-2) with v. See what you can do with this hint. Please share your work so we can suggest some hints.

 

[Otis]

for which values of [parameter a does] this [equation have] at least 1 solution

Hi GP. Did they specify Real solutions?
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[GilgameshProject]

Hi GP. Did they specify Real solutions?
[imath]\;[/imath]

Yep. Well I sure hope so, it's for 9th grade. Yes, solutions in Real numbers

 

[GilgameshProject]

Replace sqrt(x-1) with u and sqrt (a-2) with v. See what you can do with this hint. Please share your work so we can suggest some hints.

 

[GilgameshProject]

not quite sure, what to do with the 4 - 24v + v^2 here

 

[GilgameshProject]

 

[GilgameshProject]

So I did it to the point when I know which values of "a" give me discriminant = 0, but still not sure, how to check, how to check roots for being > 1(initial condition for existing of sqrt(x-1) in denominator)

 

[BigBeachBanana]

Let [imath]u = \sqrt{x-1}[/imath] and [imath]v = \sqrt{a-2}[/imath] as suggested above.
[math]\dfrac{25}{u} + \dfrac{4}{v} = 14 - u- v\\ \left(\dfrac{25}{u} + u -10\right)+ \left(\dfrac{4}{v} + v -4 \right)= 0\\ [/math]
Can you finish?

 

[Cubist]

View attachment 35065

Your work above is very good, except that a "v" disappears and then it reappears on the next line ...



Can you please explain your thinking on the following step...



I would have gone down this route...

D≥0
(v² - 24v + 4) (2-v)² ≥ 0
v² - 24v + 4 ≥0 since (2-v)² ≥ 0

 

[GilgameshProject]

Let [imath]u = \sqrt{x-1}[/imath] and [imath]v = \sqrt{a-2}[/imath] as suggested above.
[math]\dfrac{25}{u} + \dfrac{4}{v} = 14 - u- v\\ \left(\dfrac{25}{u} + u -10\right)+ \left(\dfrac{4}{v} + v -4 \right)= 0\\ [/math]
Can you finish?

Wow, yes I can! Thanks so much, don't know how I didn't see this

 

[GilgameshProject]

Thanks very much, everybody!

 

[Cubist]

I would have gone down this route...

D≥0
(v² - 24v + 4) (2-v)² ≥ 0
v² - 24v + 4 ≥0 since (2-v)² ≥ 0

If you continue with the route above, then don't forget that (2-v)² = 0 causes D=0

But, actually, the route suggested by @BigBeachBanana in post#11 is less painful.
See if you can turn it into (something squared)/u + (something squared)/v = 0 and remember that u and v are ≥ 0

Wow, yes I can! Thanks so much, don't know how I didn't see this

I see that you got there. Well done!

Spoiler

Post#11...
[math]\left(\dfrac{25}{u} + u -10\right)+ \left(\dfrac{4}{v} + v -4 \right)= 0 [/math]Then...
[math](u - 5)^2/u + (v - 2)^2/v = 0[/math]

 

[Steven G]

not quite sure, what to do with the 4 - 24v + v^2 here

If you want real solutions, then you want 4 - 24v + v^2>0

 

[Otis]

If you want real solutions, then you want 4 - 24v + v^2>0

Hi Steven. You mean 4 – 24v + v^2 > 0 AND v>0, yes?

Yet, there's more to it than that. Squaring both sides of an equation may lead to extraneous solutions, so we need to verify our solution set.

EG: Let v = 0.16

4 – 24(0.16) + (0.16)^2 = 0.1856

Therefore, we have satisfied 4 – 24v + v^2 > 0 AND v>0

Solve 0.16 = √(a – 2)

a = 2.0256

That parameter value leads to an equation with no solution:

25/√(x – 1) + 25 = 13.84 – √(x-1)

I haven't invested much work on the given exercise, but so far I've found only one Real value for parameter a that yields a Real solution for x.
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