solving a HARD SAT big exponent equation the math way

-----------------------------
If you find my channel helpful and would like to support it 💪, then you can

-----------------------------
#justalgebra #SAT

[gamingmusicandjokesandabit]

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

[Johnny-twpr]

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

[JeremyLionell]

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

[aether]

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

[mastermind]

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 :)

[miajul]

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

[佐藤広-cp]

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.

[OptimusPhillip]

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

[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!

[davidbrisbane]

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.

[gayatrim]

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

[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

[kepler]

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

[Dalton]

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

[JSSTyger]

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

[ethyl_alcohol]

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

[umeshpadwal]

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

[JakeMarley-kg]

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

[rediskathefox]

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

[hehxd]

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)