To solve this you can let a^2 be X, then you get the equation X^2 - 3 X - 4 = 0 => (X + 1)(X - 4) = 0. Then replace X with a^2 and then the solution is: a^2 = -1 => a = sqr. rt. a^2 = +/- sqr. rt. -1 = +/-i and a^2 - 4 = 0 => (a + 2)(a - 2) = 0 => a = +/- 2 (4 solutions: +/- 2 and =/- i).
Is it weird that I started on solving multi variable derivatives and integrals before really learning college algebra? I did pretty well with algebra in school so I decided to jump right into the deep end and I've been back peddling to fill in some gaps, but doing this has definitely increased my learning rate because I feel like I have this incentive to pick up these principals and run with them. If I had started with the mundane stuff like continuity, analyzing functions and graphs.. then I would not have stuck with learning maths and engineering.
yes, the explanation was awful. step 1: cubical. ax^3 + bx^2 + cx + d; becomes 1x^3 + 1x^2 + 0x - 36; a^3 + a^2 - 36; notice the last term is negative. factors of 36: 1,36,2,18.3,12,4,9. try each f(x) factor to see what = 0; 3 works a-3=0; have students do division. (a^3 + a^2 - 36)/(a-3) (a-3)(a^2 +4a +12) use the quadratic formula. 1 real and 2 imaginary.
Cubed=s were not involved so factoring to squares and numerals is all that is appropriate. So a^4 - 3a^2 - 4 = 0 factors down to (a^2 - 4)(a^2 +1) = 0. No cubes.
To solve this you can let a^2 be X, then you get the equation X^2 - 3 X - 4 = 0 => (X + 1)(X - 4) = 0. Then replace X with a^2 and then the solution is: a^2 = -1 => a = sqr. rt. a^2 = +/- sqr. rt. -1 = +/-i and a^2 - 4 = 0 => (a + 2)(a - 2) = 0 => a = +/- 2 (4 solutions: +/- 2 and =/- i).
this was high school math for me...
I don't remember how to do it, but this was high school math.
a=+i,-i,2,-2 ... which might be wrong .. I factored out (a²+1)(a²-4)=0, a²=-1 a²=4 which is ±i and ±2 ... I thnk
Right answer.
Your thinking is GOOD!
a^4 - 3a^2 - 4 = 0 => (a^2 - 4)(a^2 +1) = 0 => a^2 = 4 and a^2 = -1 => a = ±2 and a = ±1i
Is it weird that I started on solving multi variable derivatives and integrals before really learning college algebra? I did pretty well with algebra in school so I decided to jump right into the deep end and I've been back peddling to fill in some gaps, but doing this has definitely increased my learning rate because I feel like I have this incentive to pick up these principals and run with them. If I had started with the mundane stuff like continuity, analyzing functions and graphs.. then I would not have stuck with learning maths and engineering.
How can I differentiate a polynomial from a quadratic equation
(a+1a-3) (a+2a-2)
a+1a = 2a and a+2a = 3a so (a+1a-3)(a+2a-2) => (2a -3)(3a-2)
I love you ❤
a=2
or -2
9 minutes later..........
I saw the answer in 3 secs without the taliking
explain
Talk to much 😂
yes, the explanation was awful.
step 1: cubical. ax^3 + bx^2 + cx + d; becomes 1x^3 + 1x^2 + 0x - 36;
a^3 + a^2 - 36; notice the last term is negative.
factors of 36: 1,36,2,18.3,12,4,9.
try each f(x) factor to see what = 0; 3 works
a-3=0; have students do division. (a^3 + a^2 - 36)/(a-3)
(a-3)(a^2 +4a +12) use the quadratic formula.
1 real and 2 imaginary.
Cubed=s were not involved so factoring to squares and numerals is all that is appropriate. So a^4 - 3a^2 - 4 = 0 factors down to (a^2 - 4)(a^2 +1) = 0. No cubes.
Wow--way too complicated and easy to make mistakes. Most viewers will be turned off by this video and be driven away from advanced algebra.
I disagree. I struggle with maths but this was explained so well even I was able to follow it.