Evaluating A Radical Expression | Two Methods

[@denkid4478]

Let y=✓x So y^2-5/y=6 Giving y^3-6y-5=0 Overall. Can be written as y^3+1-6(y+1)=0 using y^3+1=(y+1)(y^2-y+1) and taking common terms together one gets (y+1)(y^2-y-5)=0 As y cannot be -1 (y=✓x y^2-y=5 which one recognises as x-✓x=5

[@ronbannon]

Very nice. Will share this with my students!

[@SyberMath]

Awesome! Thank you!

[@NadiehFan]

I solved it by manipulating the given equation x − 5/√x = 6 Multiplying both sides by √x we have x√x − 5 = 6√x x√x − 6√x − 5 = 0 (x√x + 1) − 6(√x + 1) = 0 (√x + 1)(x − √x + 1 − 6) = 0 Since √x + 1 ≠ 0 this implies x − √x + 1 − 6 = 0, that is, x − √x = 5. The converse derivation is a lot simpler: if we are given that x − √x = 5 then dividing both sides by √x gives √x − 1 = 5/√x and therefore x − 5/√x = x − (√x − 1) = x − √x + 1 = 5 + 1 = 6.

[@SyberMath]

Nice

[@scottleung9587]

Nice!

[@SyberMath]

Thanks!

[@giuseppemalaguti435]

La prima equazioni è una cubica con soluzioni √x=-1(non accettabile)..√x=(1+√21)/2 (accettabile)...√x=(1-√21)/2(non accettabile)...perciò ?=(11+√21)/2-(1+√21)/2=5

[@uwelinzbauer3973]

I came to the cubic equation. Trying to plug in x1=1 worked, then I divided the polynomial by (x-1), got a quadratic equation, delivering x2,3=(11±Sqr(21))/2 Now we arrive at the discussion: are we allowed to give root of x a negative sign? If yes, then we have three solutions for x. And that would mean six different answers to x-sqr(x). If no, only one solution for x: x=(11+sqr(21))/2≈7.79 and x-sqr(x)=5 Thanks for the interesting videos, Greetings.

[@uwelinzbauer3973]

Important attachment to above / error correction: x-sqr(x) does not have 6 different answers, a further check showed that there are only 5 different answers, because the value 5 is appearing twice Sorry, I have to apologize for the mistake. - BTW: is (-1) a root of 1? To my opinion it might be, what do you think? If no: why?

[@N79871]

Nice

[@SyberMath]

Thanks

[@mohammadazadi4535]

با در نظر گرفتن -x_x^1/2بعنوان جمله u وضرب مزدوج آن در دوسوی معادله اول جمع دوطرف معادله اول با آن و... جواب بدست آمد مسله جالبی بود.

[@kianmath71]

5

[@rakenzarnsworld2]

Answer: 5

[@mcwulf25]

In your first cubic, x=1 is a solution. However, it doesn't work in the original equation because it arises from a negative square root.

[@bobbyheffley4955]

Extraneous solution

[@chaosredefined3834]

Before watching the video: Let t = sqrt(x) Therefore, we have t^2 - 5/t = 6, and we want to find y = t^2 - t t^2 - 5/t = 6 t^3 - 5 = 6t t^3 - 6t = 5 Now, I had a plan here, but then I noticed that t = -1 is a root of that equation. t^3 - 6t - 5 = 0 t^3 + t^2 - t^2 - t - 5t - 5 = 0 t^2(t+1) - t(t+1) - 5(t+1) = 0 (t^2 - t - 5)(t + 1) = 0 t^2 - t - 5 = 0 t^2 - t = 5 y = 5 OR t+1 = 0 t = -1 t^2 = 1 t^2 - t = 1 - (-1) = 2 So, y = 2 or y = 5.

[@chaosredefined3834]

wait. t = -1 is a superfluous root. So, y=5 is the only solution.

[@-basicmaths862]

Actually 2 is only real answer. √x=-1 satisfied above equation.if x=1 then √x=±1