Solving a Parametric Equation

[@Relrax]

due to x >= 1, the simple substitution x = k^2+1 also solves the problem rather quickly.

[@SyberMath]

👍

[@chaosredefined3834]

After watching enough of your videos, I decided to employ your most powerful trick: Substitution! x + sqrt(x - 1) = a Let u = sqrt(x-1) u^2 = x - 1 u^2 + 1 = x Therefore u^2 + u + 1 = a u^2 + u + (1-a) = 0 Quadratic formula gives us: (-1 +/- sqrt(1 - 4(1-a)))/2 = u -0.5 +/- sqrt(a - 0.75) = u Since u is defined as sqrt(x-1), it must be positive. As -0.5 is negative, that means we can't use one of the two solutions. So we are left with: -0.5 + sqrt(a - 0.75) = u Squaring that gives us: 0.25 + sqrt(a - 0.75) + a - 0.75 = u^2 a - 0.5 + sqrt(a - 0.75) = u^2 a + 0.5 + sqrt(a - 0.75) = u^2 + 1 = x I'd normally avoid the decimals, but trying to write that out in text format was annoying. But that substitution makes the quadratic formula stuff much easier. The rest of the process is the same, except we took care of one of the solutions earlier.

[@SyberMath]

Pretty good!

[@aashsyed1277]

@@SyberMath (!@#$%^&*(+-*/-+

[@thomasstokes9412]

If we are just looking for integer solutions then (x,a)={(k^2-k+1,k^2-2k+2)|k\in N} generates all the integer solutions. This is derived by letting 4a-3=(2k-1)^2 in the quadratic equation.

[@SyberMath]

Nice!

[@SyberMath]

This is an equation with parameters. Parametric Equations are fairly interesting. They are harder but more fun to solve. Any thoughts? Here are some similar videos: Solving a radical equation with a parameter: kzbin.info/www/bejne/g3-lqpmeibqiq6M Solving a cubic equation with parameters: kzbin.info/www/bejne/hnnRh3h3qb6niLM Solving a cubic equation with a parameter: kzbin.info/www/bejne/Y4qskJd7e6hmf6s Solving a quartic equation with parameters: kzbin.info/www/bejne/fKaWoqp5rJd4qtk

[@זאבגלברד]

The left side is an increasing function starting at (1,1). So for every a>=1 there is exectly one solution. So to select the one solution in the quadratic formula, we can take a=7 that comes from x=5 and select the correct one.

[@SyberMath]

Nice!

[@2false637]

Really nice

[@SyberMath]

Thanks a lot

[@Alians0108]

My method: x + √(x-1) = a Minumum of the function on LHS, as it is strictly growing, is when x = 1 -> 1 + 0 = 1. so a => 1 x-1 + √(x-1) = a-1 (x-1) + √(x-1) + (1-a) = 0 Quadratic in terms of √(x-1) √(x-1) = -1/2 ± √(1/4-1+a) = -1/2 ±√(a-3/4) √(x-1) = (-1± √(4a-3))/2 (x =>1) -1 - √.... < 0 (negative root is false) √(x-1) = (-1+√(4a-3))/2 x-1 = (1-2√(4a-3)+4a-3)/4 = x = 1 + (-2-2√(4a-3)+4a)/4 = a + (1-√(4a-3))/2 So: x = (2a + 1-√(4a-3))/2 if a => 1

[@SyberMath]

Pretty

[@souhilaoughlis5832]

I always enjoy your videos

[@SyberMath]

Thank you!

[@Артьомдругартем]

Можно было найти экстремум функции в левой части. (x+sqrt(x-1))'=1+1/2*sqrt(x-1)>0. Функция возрастает(это и без производной понятно). Min f(x)=f(1)=1

[@davidseed2939]

Should you also check the solution satisfies x

[@SyberMath]

Didn't we check that?

[@leif1075]

@@SyberMath you meant to say x has to be.greeater.than ot equal to 1 right? At 1:04?

[@SyberMath]

@@leif1075 I said x≥1, didn't I?

[@akakidzidziguri7947]

1) x=1 a=1 2) x=2 a=3 I don't know if there is more solutions. Waiting for video👌❤

[@srijanbhowmick9570]

Umm , the value of "x" should be in terms of "a"

[@SyberMath]

Yep yep!

[@akakidzidziguri7947]

I thought we wanted only positive integers😁 I also did same that was in video, but anyways it was nice video.❤

[@SyberMath]

@@akakidzidziguri7947 Thanks!

[@joaquingutierrez3072]

Nice video !!

[@SyberMath]

Thanks!

[@seshnarayan7972]

2:16😂😂😂

[@SyberMath]

😂

[@aashsyed1277]

NYC!

[@KingGisInDaHouse]

What you wrote was minus or plus not plus or minus. Doesnt make a difference here but there are instances where you would have to use both and it indicates order.

[@SyberMath]

Yeah, I always write it this way!

[@ilyaafminskilyaafminsk6268]

А можно было и так: x-1+✓(x-1)=a-1 1

[@SyberMath]

Very nice!

[@srijanbhowmick9570]

Hey SyberMath , is my answer correct ? (Press Read More) x = ((2a + 1) - sqrt(4a - 3))/2 where a >= 1

[@SyberMath]

☺️

[@srijanbhowmick9570]

@@SyberMath Umm , I don't understand what you mean

[@SyberMath]

@@srijanbhowmick9570 looks good!

[@srijanbhowmick9570]

@@SyberMath 😁

[@deepjyoti5610]

Wow nyccc"ccccccc×100000000 vedio

[@seshnarayan7972]

2:16😂😂😂

[@SyberMath]

😂