solving a HARD SAT big exponent equation the math way

Solving a HARD SAT big exponent question the math way! We are given x^3(x^2-5)=-4x and we have to just find one answer. However, after distributing the x^3 and rearranging the terms, this is actually a solvable quintic equation (a polynomial equation with degree 5). We can solve all the solutions by factoring! But if you want a shortcut, then check out • Shorts
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"Just Algebra" (by blackpenredpen) is dedicated to helping middle school, high school, and community college students who need to learn algebra. Topics include how to solve various equations (linear equations, quadratic equations, square root equations, rational equations, exponential equations, logarithmic equations, and more), factoring techniques, word problems, functions, graphs, Pythagorean Theorem, and more. We will also cover standardized test problems such as the SAT. Free feel to leave your questions in the comment! Subscribe for future videos 👉 bit.ly/just_al...
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#justalgebra #SAT

[@Johnny-tw5pr]

If this is a HARD SAT question then I'm confident I can get a really good score

[@mananagrawal7458]

boy oh boy, sat is a standard test if you really wanna test your skills then give the hardest high school exams in the world like jee advance

[@darkrising8280]

Me who has horrible algebra foundations becuase of covid 🙂

[@zealous919]

How wtf? I’ve seen analysis proofs easier to understand than this. I had no idea you could factor that quartic function like that

[@darkrising8280]

@@zealous919 that's becuase he never tried the test

[@spooky2526]

@@darkrising8280 sorry but no, the sat is incredibly easy by international standards, I've been doing this type of double quadratics in a standard math class since the last year of middle school

[@gamingmusicandjokesandabit1240]

2:50 Steve: *uploads to just algebra* Also Steve: *draws an integral sign to keep writing*

[@aether4167]

Can it also be, consider x^4-5x^2+4 is z^2-5z+4 and now you can use quadratic formula

[@Pankaw]

Yup, but if you can, factorize it

[@aether4167]

@@Pankaw i see, i will try it out

[@scurfle938]

The quadratic formula only accounts for 2 out of 5 total solutions, 3 with the coefficient of x on the left. So, you have to factor to find the remaining solutions (or use the endlessly long quartic formula that would take literal hours to do by hand :)

[@Pankaw]

@@scurfle938 Well, you would get that x² = (-b ± sqrt(b² - 4ac)) / 2a Then x = ± sqrt((-b ± sqrt(b² - 4ac)) / 2a) Which are four solutions The fifth solution comes from the factor that was left outside (just x)

[@aether4167]

@@Pankaw thanks

[@JeremyLionell5]

2:13 I think there's a better sol: x⁴-5x²+4=0 (x²)²-5x²+4=0 Let x²=y y²-5y+4=0 Factor it (y-4)(y-1)=0 y=4 or y=1 x²=4 or x²=1 sqrt both sides x=±2 or x=±1

[@user-wu9hy4lt2w]

This looks complicated, but it's good to say that it was actually just the product of a simple primary and secondary equations. In the case of x=0, the left side = 0^3 (0^2-5) = 0, the right side = -4*0=0, the left side = the right side, it is one of the solutions, so in the case of x≠0, if both sides are divided by x, the solving may be advanced as x^2 (x^2-5)=-4.

[@miajul867]

I think that a better way for solving the equation after the development + factorisation by x This: x(x^4-5x^2+4)=0 Is to create a new variable X=x^2 for the second part Then X^2-5X+4=0 And then use delta and transform X into x

[@mastermind1258]

In German schools we learn to solve this type of question like this: 1. Do all the steps like you did, till factoring the x out 2.Factor the x out 3. Substitute x² with a variable like "a" 4. Former step gave you a quadratic formula you can solve with the abc-formula 5. Resubstitute a with x² -> you get just the same values. With this technique I solved this problem in literally under one minute :)

[@TPColor]

yeah i did it this way

[@benberk8541]

I guess they want us to get lots of practice factoring so they tell us to use the longer way.

[@oZqdiac]

You don’t even have to use the quadratic formula when you get your quadratic equation in this case It is factorisable

[@user-xv7xq3wt4x]

@mastermind1258 I solved it about as fast, but my method was to just solve it using a method similar to how I learned to solve cubic equations.

[@PragmaticAntithesis]

Firstly, all terms contain x so x=0 is a trivial solution. For the other solutions: Divide by x^3 (x=0 is already accounted for) x^2 - 5 = -4x^-2 Let y=x^2 y - 5 = -4/y Multiply by y y^2 - 5y + 4 = 0 Factorise (y-4)(y-1) = 0 Divide by each bracket y = 4 or 1 Take square roots for x x = any of (-2, -1, 0, 1, 2) How cute, the solutions are the five smallest integers!

[@theonewiththegoldentouch]

This guy finds math cute

[@gayatrim1615]

I got x= (+/-)1 , (+/-)2 by keeping x^2 as a variable like "a" and solved the quadratic equation

[@maoitsme0]

same

[@OptimusPhillip]

This is essentially a quadratic of x^2, with an additional solution of 0. -2, -1, 0, 1, and 2 are all solutions

[@Dalton1294]

I did it in my head and got x=-2, x=-1, x=0, x=1, and x=2

[@deltalima6703]

I did it in my head and missed x=0. Meh, dont care, I am not in math class, lol. I got the other 4 and double checked them in my head too. I subbed y = x squared in my head and did it that way.

[@ethyl_alcohol774]

This video and the comment section is very helpful! I solved this equation using synthetic division so it was very long and time consuming 😂

[@user-xv7xq3wt4x]

As a rising 9th grader who solved this in a minute or two, i think i am going to do good on the SAT.

[@davidbrisbane7206]

Consider the equation x⁴ - 5x² + 4 = 0 and find its factors. Applying the _Rational Root theorem_ then the possible roots are -1, 1, -2, or 2. It turns out that they are all roots. So we have found the remaining four roots of the quintic equation, with the last root being x = 0. Problem solved. Note: This approach does not always work, but it works often enough in exam questions where the numbers involved are often just integers or rational numbers.

[@Acropolis_]

I think we can change x3 to x2 and 4x to 4 by dividing by x then you can do a 2nd degree polynomial by x2 = z

[@wzshad8031]

Great teacher

[@rediskathefox]

1:57 i imagine how i will do it. I think it is illegal. But, actually cool! 😂

[@derkion6436]

Sorry but this is basic math in turkey if i see this question i will be happy because its cheap point for me. Please try our universty exam questions(yks-ayt) they are ruthless

[@mustakbelmuhendis4172]

Kanka napıyon

[@JSSTyger]

I came up with 5 solutions...0, -1,1,-2, and 2.

[@tk1699]

yes that is correct

[@createyourownfuture5410]

But how do we know that those are all the solutions?

[@JSSTyger]

@@createyourownfuture5410 It can be broken down to x(x^4-5x^2+4) = 0. This automatically means that x = 0 is one solution. If we let x^2 = a, we get x(a^2-5a+4) = 0 and you can get (2) solutions for "a", resulting in (4) more solutions for x.

[@createyourownfuture5410]

@@JSSTyger Yes, I solved it by doing exactly that, and 5 solutions for a quintic equation is confirmed by the fundamental theorem of algebra, but is that all in the complex world too? I think the answer is yes, as the FTA accounts for complex numbers as well, but I would like if someone confirmed it.

[@affapple3214]

@@createyourownfuture5410 Yes the answer also accounts for the complex solutions. As the fundamental theorem of algebra says,there is 5 solutions. So if you found 5 solutions and they are all correct, there is no complex root to this equation. All the solutions are real

[@kepler4192]

yes I did pause this time, and I ended up with 5 solutions: x=0, x=-1, x=+1, x=-2, x=+2

[@bruhbro1181]

The solutions are all integers from -2 to 2 inclusive

[@kepler4192]

@@bruhbro1181 that's actually a cool thing about this equation

[@bruhbro1181]

@@kepler4192 so u can simplify x = all integers between -2 and 2 inclusive

[@hehxd1042]

I solved it this way: x^3(x^2-5)=-4x First of all, it is obvious that x no.1 is zero. x^5-5x^3=-4x Dividing both sides by x, we get x^4-5x^2=-4 x^4-5x^2+4=0 Let’s substitue u, where u=x^2 Therefore, (u-4)(u-1)=0 So, x no.2,3=+ or -(2) And x no.4,5=+ or -(1)

[@tbg-brawlstars]

I did in 15-20 seconds in my mind 😄

[@alibekturashev6251]

You can write the answer as ±(1±1)

[@karllytskfk8471]

I read this as {-2;2;0}. There's no 1. Don't know if a simple notation exists

[@Shivshankar_JEE-ov5wx]

The way he solved q.eqⁿ forming boxes was amaze ♥️

[@atifyt2751]

yeah hpw to do that ???

[@umeshpadwal5823]

It is so easy you can simple use quadratic formula. We can put X = 1 also

[@Vladimir_Pavlov]

".....solving a HARD SAT big exponent equation ....." ?????????????

[@tumak1]

...set u=x^2 at one point and it is so much easier to solve. Cheers

[@zainabhusain4076]

This is really good equation

[@Nothingx303]

I have solved this question by a different method Divide both sides by x³ so we get X² -5 =-4/x² Let x² =y Y-5 = 4/y After solving we get Y = 1 or 4 So now we get x = ±1 and ±2 But 😂 x=0 left my solution 😊

[@justamir1558]

Exactly I did the same and I couldn't find zero too.

[@Brid727]

this is so easy 😂 Just know when x is a multiplicand and/or multiplier on both sides one of the solutions is always 0. Cancel x from both sides so that it becomes x^2(x^2 - 5) = -4 Distribute L.H.S x^4 - 5x^2 = -4 x^4 - 5x^2 + 4 = 0 Let x^2 = a a^2 - 5a + 4 = 0 (a - 1)(a - 4) = 0 a = 1, a = 4 x^2 = 1, x^2 = 4 So, x = (-2 , -1 , 0 , 1 , 2)

[@necaro]

Ruffini´s rule would also work

[@tamirerez2547]

Hello, Mr. Teacher! First, thank you for good video, and simple nice solution. One small question always bothers me. We are always learning how to solve such a problem. It's OK. But what I don't understand is, how do you invent such a problem?? Who makes up these questions? What is the technique? How do you invent a math problem that will surely be solvable? And it will have whole number answers. Who invents this??? Can you show us a video where you, the teacher, come up with a math problem, show us how do you"invent" such a problem with clear whole number solutions? How do you know that it can be solved, etc., etc. Will you accept this challenge?

[@mrgamer53082]

I solved this very easily

[@tanishdesai7652]

I could solve it. Thanks, nice vids btw

[@juancarlosnadermora716]

Got it! 5 solutions in the reals!

[@juandelacruz9027]

Nice. I got the same answers.

[@xSayf_Rx]

The fact that he held the pokemon ball the whole way through made it thousands time better 😂

[@hezareax]

its the mic

[@kingmystery2136]

Solved it in seconds

[@devondevon4366]

x= -1 , 1 , 2, -2 and 0

[@aryansinha4992]

....I legit just guessed all five of the roots

[@user-yg9zb4qi2g]

Good 👍

[@rkrishnamoorthy1785]

X is just unity.

[@dotcomgamingd5564]

honestly that was super easy once you learn how to break it down

[@lyf1358]

I found 0 ; +-i ; +-2i is that correct ?

[@oZqdiac]

No there are no complex solutions

[@jai_chauhan_]

I'm an indian and am in grade 9 and I found it extremely easy to solve

[@anestismoutafidis4575]

X=2

[@oZqdiac]

What about the other 4 solutions

[@yasith2426]

What is the SAT and is this really a "hard" question?? Seems very straightforward

[@anshumanagrawal346]

Yeah, I think it's pretty a simple question

[@TPColor]

Yeah its not that hard once you know how to solve these kinds of equations. The hardest math questions on the sat, at least for me, are the word problems and maybe a system of 3 polynomials

[@deltalima6703]

I think SAT is american high school, but I dont know for sure, I am not american.

[@ZipplyZane]

The SAT is essentially a test that some Americans take in high school as part of qualifying for college. The math doesn't go up past trigonometry, if I remember correctly. The ACT, a similar test, definitely doesn't go past trig. I don't think it's _actually_ hard. He's just saying it looks hard for someone in an algebra class. If you're taking regular algebra in college (i.e. the class he is teaching), you probably struggled with math in high school to some degree.

[@waynelw4141]

Can I divide both side by X at the start? Found all the answer except x=0

[@oenrn]

You can't divide by zero, so if you divide by x you exclude x=0 as an answer.

[@itsSujeetMahto]

Solved under 10 seconds, what kinda test is it?

[@boxman_ninja0819]

1

[@madlad8521]

Ayo if this is hard ima score a good score (cbse 10th btw

[@gayatrim1615]

Lol same here...i looked at the thumbnail and immediately solved it

[@lolzazo]

meanwhile me who got x = i*sqrt(0.8) ...

[@hans429]

From the looks it seems like -1 could work...

[@MdJamal-cp4zl]

I have gotten 2 answers only

[@abdussamad01736]

I found only two answers 2 and-2.

[@fuckingdumbo]

dude no way that's an SAT question It's like a beginner level problem

[@fuckingdumbo]

@@bprpmathbasics yeah sure thanks

[@rom5457]

Lol tried with 2

[@darkrising8280]

To the majority of people that say it's easy. Have you took the test or at least a practice test 🤨

[@GirishManjunathMusic]

x³(x² - 5) = - 4x x³(x² - 5) + 4x = 0 x(x²(x² - 5) + 4) = 0 x(x⁴ - 5x² + 4) = 0 x(x⁴ - x² - 4x² + 4) = 0 x(x²(x² - 1) - 4(x² - 1) = 0 x(x² - 4)(x² - 1) = 0 x(x - 2)(x + 2)(x - 1)(x + 1) = 0 x ∈ {-2, -1, 0, 1, 2}

[@A_Random_Melon_Pult]

this video popped up on the recommend videos and as a 7th grader i dont really get it...

[@ritiknsutdelhi122]

I dont even lift the pen for this one grow up man.

[@themathsgeek8528]

This was easy..

[@eduardokuri1983]

So hard SAT=Kinda tricky

[@vexor4083]

So many dunning-Krugers in these comment sections

[@amirmahadeo56]

This was a bit too easy

[@ruchikarfacts7380]

Too hard to solve.😂

[@KiranYadav-qp4sp]

it is not hard at all.

[@ZainAli-zv8nt]

I am feelin sleepy. This is hard? Who the hell are you fooling?

[@tbg-brawlstars]

Before seeing the video, I am writing this comment so to see if my answer is right or not x=0,±1,±2 Edit : I was right :) If I would be wrong, I would be ashamed to not solve this simple problem....