Solve (a to 4th – 3a squared – 4 = 0) How Much Algebra Do You Know? Polynomial Equations

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[russelllomando]

WOW. once i saw the substitution...i solved it in 1 minute. great idea. thanks for the fun

[mercy]

We have a⁴ - 3a² - 4 = 0 — we gonna use the “u-substitution” method


1. We’ll break down the a⁴, so we have:
(a²)² - 3a² - 4 = 0

2. Since there’s 2 a²’s, we can use u-substitution to get:
Let u = a²

u² - 3u - 4 = 0

3. This becomes a quadratic equation where we can easily factor and solve for u:

(u - 4)(u + 1) = 0

4. Using zero product property, we can determine that:

u - 4 = 0, u + 1 = 0 u = -1, 4

5. Now, we’re not done, remember that we let u = a², so we’re going to plug in -1 and 4:

a² = -1 a² = 4
a = ±√(-1) a = ±√(4)
a = ±i a = ±2

Final ans: a = ±2, ±i

Ez

[johnplong]

This is a Algebra 2 However in my high school This was called math 4 which was taken after Algebra 2 I got introduced to this at the end of Algebra 2 We learned about this after learning Logarithms We didn’t have calculators back in the early 70s

[johnlakin]

Thanks you tube math man, that was one of the best videos you’ve ever done😊

[Kleermaker]

We see that factoring the 'normal' way does not help us. We also see that the equation a^2 - 3a - 4 = 0 can be factored easily: (a - 4)(a + 1). So we have to replace a^2 with x (or any other letter). Then we get the same result, but have to take the square root of the solutions (4 and 1) to get the solution for a, because we said let x be a^2. So a^2 = 4 and a = square root a^2 = square root 4 = 2 and - 2. If we plug those numbers in the original function we get 2^4 3(2^2) - 4 = 0, which is correct (16 - 12 - 4 = 0) and (-2)^4 - 3(- 2^2) - 4 = 0, which is also correct (16 - 12 - 4 = 0. So now we have 2 solutions, but there are 4 solutions (an equation with a^4 has 4 solutions). So we also have to find the square root of a^2 = - 1, and square root a^2 = square root -1, which gives us two imaginary numbers: i and - i. Final conclusion: the solution is: a = 2, - 2, i and - i.

[stevenjohnson]

All I was treated this as a quadratic breaking it into (a^2-4)(a^2+1)=0 this gets (a^2-4)=0(a^2+1=0 solving the each new equation you get a^2=4 taking the square root give a=-2 and a=2 solving the other a^2=-1 taking that square root get a=-i a=i

[hgu]

set t=a^2 => t^2-3t-4=0

via quadratic formula you get t=(3+-i√7)/2

since t=a^2, a=+-√t

a=+-√((3+-i√7)/2)

[johngaffney]

I'm impressed with the knowledge you have, I've been trying to get back into mathematics, for fun, as well as increase my memory through healthy diet and exercise, still trying to see if there's a Nutropic that actually works...

[devonwilson]

Greetings. If we write
A^4-3A^2-4 as
A^4-3A^2=4, then
A^2(A^2-3)=4, and A^2=4,
(A^2)^1/2=4^1/2,
A=2, -2 for two of the roots.

[sho-nuff]

Hi. Honest question here. What do you solve using equations like this? What are the applications?

[prollysine]

let a^2=u, u^2=a^4, u^2-3u-4=0, u=(3 |+/-| V(9+16))/2, u=(3 +/- 5)/2, u= 4, -1, a^2= 4, -1, a= +/- V(2), +/- V(-1), a= 2, -2, i, -i,

[tomtke]

a^4-3a^2-4=0

I'm seasoned to solve stuff like
B^2 - 3B - 4 = 0
So what if I assign B = a^2?

B^2 - 3B - 4 = 0
(B )(B ) = 0
by inspection
-4 = -4×1
-3 = -4+1
(B - 4)(B + 1) = 0

replace
B = a^2 yields

(a^2 - 4)(a^2 + 1) = 0
first solutions
a^2 = 4
a = +/-2 sol.1, sol.2
second solutions
a^2 = -1
a = +/- i sol.3, sol.4

VERIFY
a^4-3a^2-4=0

a=2
(2)^4-3(2)^2-4=?0
16-12-4=?0
0=❤0✔️

a=-2
(-2)^4-3(-2)^2-4=?0
16-12-4=❤0✔️

a=i
(i)^4-3(i)^2-4=?0
1-3(-1)-4=?0
1+3-4=?0
0=❤0✔️

a=-i
(-i)^4-3(-i)^2-4=?0
(-1)(-1)-3(-1)-4=?0
1+3-4=❤0✔️

[chrisdissanayake]

a^4 - 3a^2 - 4 = 0

Factoring the above,

(a^2-4) (a^2+1)=0

a^2-4 =0 or a^2+1=0

If a^2-4=0,
a^2=4
a=+2 or -2

If a^2+1 =0,
a^2=-1
a= +square root of (-1)
OR
a= -square root of (-1)

Square root of (-1) = i

Therefore,
a= +2 or -2 or +i or -i

[walterwen]

(a^2 + 1)(a^2 - 4) = 0
a^2 = - 1; a = +/- I
a^2 = 4; a = +/- 2

[jimwetzel]

I kind of like a = +/-2 or a = +/- sqrt(-1), better known as "i"

[dwaipayandattaroy]

Perhaps it's a^4+ 3a^3 -4=0
Ie a^2( a^4+ 3a -4), cause all a^2 consumed in common taking, then 3a^2 left with 3 only,
And why they kept -4 uneffected from a^2 common taken effect, chill this much you have in memory that doesn't get effected,

But
a^4+3a^3-4
Comes as
a^2- 3a -4 that is doubtful, if a^2 taken from 3a^2 then only 3 is left, isn't it, maths some people believe that 2+2=/ 4, it's 5 instead 💀😂✌