It's been so long since I took an official algebra class. I got to (x^2+4) and wondered what to do next. And then I realized I needed it to equal zero. Thanks so much for allowing us older people to start exercising our math brains in an entertaining way!
I have watched a lot of your videos and many of them are just reviews of stuff I have forgotten from when i was in high school 50 years ago. This one is as far as I can tell is completely new to me, and I'll admit that it hasn't quite sunken in. Great job of teaching math John!
I did use the old school method: group factoring. X^2(X - 2) + 4(X - 2) = 0 => (X^2 + 4)(X - 2) = 0. So the solutions are: X = 2, that gives 8 - 8 + 8 - 8 = 0 (correct) and (X^2 + 4) = 0 => X^2 has to be - 4, which never is the case for any rational number, so the answer must be X = sqr. rt. - 4 which gives us +/- 2i, two imaginary numbers (I am not fond of these imaginary numbers if they stay what they are).
x=2 is clearly a solution, by three-second inspection. So x-2 factors right out: x^3 - 2*x^2 + 4*x - 8 = 0 (x-2)*(x^2 - 4) = 0 (x-2)*(x-2)*(x+2) (x+2)*(x-2)^2 So we have x = 2, x = 2 again, and x = -2 as the three solutions.
There is a general formula for third order polynomials (as well as 4th order ones). It isn't as simple as the quadratic formula (it has a lots of steps), but it does exist. With real coefficients, there is at least one real solution, and the two other roots are conjugate pairs, or both real numbers.
@@oahuhawaii2141 I believe that imaginary numbers were invented to make the cubic formula a general one. Before then you had to decide one of several conditions to pick. Who knows, one of the lessons might be the derivation of the formula. I haven't seen it derived. I did see it in a book many many years ago.
Just for grins I tried the 2i root with the synthetic division and derived the quadratic equation where I obtained the other roots. It was a bit more tedious, but it worked out.
Good problem for employing the rational roots technique for finding the first root, then using the depressed equation to find the additional roots. The fact that the additional roots were complex made the problem that much more encompassing.
x³ - 2*x² + 4*x - 8 = 0 This is a cubic equation with real coefficients, which means there are 3 roots with 1 or 3 real roots. All 3 potential answers show 1 real root and 2 complex conjugate roots. It's easy to test the real solutions of the first 2 answers listed, x = 1 or 2, so let's do that: x = 1: 1 - 2 + 4 - 8 =?= 0 -5 ≠ 0 x = 2: 8 - 8 + 8 - 8 =?= 0 0 = 0 Thus, we know x = 2 is one root, and the answer must be "b". As a check, we can find the remaining 2 roots by factoring out (x-2) from the equation: x²*(x-2) + 4*(x-2) = 0 (x²+4)*(x-2) = 0 Now, we know (x²+4) = 0, which means x = ±2*i . This matches the other values in "b". Of course, if you had noticed that the coefficients are increasing powers of 2 with alternating signs, that would immediately lead you to factor the equation after rewriting it as: 2⁰*x³ - 2¹*x² + 2²*x¹ - 2³*x⁰ = 0
You DON'T need to go through the trouble of the RRT and all that synthetic division. X^3-2x^2+4x-8=0 is factorable by grouping: X^2(x-2)+4(x-2)=0 = (x^2+4)(x-2)=0 from that you get x^2=-4: x= + or -2i and x=2. Done
I got the answer because the basic arithmetics (not guessing) told me that only B) worked. But now I really need to watch you solve this one step by step. ;)
That 2nd step isn't a good idea. Do this instead: x³ - 2*x² + 4*x - 8 = 0 This is a cubic equation with real coefficients, which means there are 3 roots with 1 or 3 real roots. All 3 potential answers show 1 real root and 2 complex conjugate roots. It's easy to test the real solutions of the first 2 answers listed, x = 1 or 2, so let's do that: x = 1: 1 - 2 + 4 - 8 =?= 0 -5 ≠ 0 x = 2: 8 - 8 + 8 - 8 =?= 0 0 = 0 Thus, we know x = 2 is one root, and the answer must be "b". As a check, we can find the remaining 2 roots by factoring out (x-2) from the equation: x²*(x-2) + 4*(x-2) = 0 (x²+4)*(x-2) = 0 Now, we know (x²+4) = 0, which means x = ±2*i . This matches the other values in "b". Of course, if you had noticed that the coefficients are increasing powers of 2 with alternating signs, that would immediately lead you to factor the equation after rewriting it as: 2⁰*x³ - 2¹*x² + 2²*x¹ - 2³*x⁰ = 0
I wouldn't even know where to start with a cubic involving imaginary numbers. I'm just chuffed I even know what any of that is and that's as far as it goes. Edit- All I remember is complex numbers are used for oscillations in quantum mechanics because i=i, i2=-1, i3=-i, i4=1,i5=i, etc
Yes you can: x³ - 2*x² + 4*x - 8 = 0 x²*(x-2) + 4*(x-2) = 0 (x²+4)*(x-2) = 0 Now, we know: (x²+4) = 0 and (x-2) = 0 x = ±2*i and x = 2 These values match the values in "b". Of course, if you had noticed that the coefficients are increasing powers of 2 with alternating signs, that would immediately lead you to factor the equation after rewriting it as: 2⁰*x³ - 2¹*x² + 2²*x¹ - 2³*x⁰ = 0
My hair are, indeed, on fire. But the title is right and my algebra knowledge is not yet that deep. But I learned a lot of new stuff with this demonstration.
Not a terribly difficult problem to pick out the correct roots. Just solve using the real number and them use the imaginary roots if you like a little more difficult challenge. B is the answer.
x^3-2x^2+4x-8=0 x^2(x-2)+4(x-2)=0 (x^2+4)(x-2)=0 I'm gonna cheat... the two solutions are x^2=-4 and x=2 x^2 yields 2 solutions for a total of 3. only option b) has x=2 as solution... in like manner.... just plug/chug with first 1 and then 5 isn't that involved.
It's been so long since I took an official algebra class. I got to (x^2+4) and wondered what to do next. And then I realized I needed it to equal zero. Thanks so much for allowing us older people to start exercising our math brains in an entertaining way!
I have watched a lot of your videos and many of them are just reviews of stuff I have forgotten from when i was in high school 50 years ago.
This one is as far as I can tell is completely new to me, and I'll admit that it hasn't quite sunken in.
Great job of teaching math John!
I did use the old school method: group factoring. X^2(X - 2) + 4(X - 2) = 0 => (X^2 + 4)(X - 2) = 0. So the solutions are: X = 2, that gives 8 - 8 + 8 - 8 = 0 (correct) and (X^2 + 4) = 0 => X^2 has to be - 4, which never is the case for any rational number, so the answer must be X = sqr. rt. - 4 which gives us +/- 2i, two imaginary numbers (I am not fond of these imaginary numbers if they stay what they are).
x=2 is clearly a solution, by three-second inspection. So x-2 factors right out:
x^3 - 2*x^2 + 4*x - 8 = 0
(x-2)*(x^2 - 4) = 0
(x-2)*(x-2)*(x+2)
(x+2)*(x-2)^2
So we have x = 2, x = 2 again, and x = -2 as the three solutions.
There is a general formula for third order polynomials (as well as 4th order ones). It isn't as simple as the quadratic formula (it has a lots of steps), but it does exist. With real coefficients, there is at least one real solution, and the two other roots are conjugate pairs, or both real numbers.
Yep, I had to derive the cubic once, but not the quartic. It was easier to look at the CRC Tables than to do all that work.
@@oahuhawaii2141 I believe that imaginary numbers were invented to make the cubic formula a general one. Before then you had to decide one of several conditions to pick. Who knows, one of the lessons might be the derivation of the formula. I haven't seen it derived. I did see it in a book many many years ago.
Just for grins I tried the 2i root with the synthetic division and derived the quadratic equation where I obtained the other roots. It was a bit more tedious, but it worked out.
Good problem for employing the rational roots technique for finding the first root, then using the depressed equation to find the additional roots. The fact that the additional roots were complex made the problem that much more encompassing.
The entire problem is factorable by grouping. No need for RRT to find the 1st root
I could see the 2 as X immediately but didn't understand the +-2i in the answer selection so I guessed b as the solution.
X=2 is a solution therefore you get. X^3 -2x^2 +4x -8= (x-2) ( x,^2 +4)=0. X= +- 2i
x³ - 2*x² + 4*x - 8 = 0
This is a cubic equation with real coefficients, which means there are 3 roots with 1 or 3 real roots. All 3 potential answers show 1 real root and 2 complex conjugate roots. It's easy to test the real solutions of the first 2 answers listed, x = 1 or 2, so let's do that:
x = 1:
1 - 2 + 4 - 8 =?= 0
-5 ≠ 0
x = 2:
8 - 8 + 8 - 8 =?= 0
0 = 0
Thus, we know x = 2 is one root, and the answer must be "b". As a check, we can find the remaining 2 roots by factoring out (x-2) from the equation:
x²*(x-2) + 4*(x-2) = 0
(x²+4)*(x-2) = 0
Now, we know (x²+4) = 0, which means x = ±2*i .
This matches the other values in "b".
Of course, if you had noticed that the coefficients are increasing powers of 2 with alternating signs, that would immediately lead you to factor the equation after rewriting it as:
2⁰*x³ - 2¹*x² + 2²*x¹ - 2³*x⁰ = 0
You DON'T need to go through the trouble of the RRT and all that synthetic division. X^3-2x^2+4x-8=0 is factorable by grouping: X^2(x-2)+4(x-2)=0 = (x^2+4)(x-2)=0 from that you get x^2=-4: x= + or -2i and x=2. Done
got it B once the 2 worked, it's solved. thanks for the fun.
I got the answer because the basic arithmetics (not guessing) told me that only B) worked.
But now I really need to watch you solve this one step by step. ;)
b) 2, +/-2i
Make video on algebra solve way
(X^3+2X^2+4X-8 =0)
X(X^2 +2X+4)=8 , then what , do to a video on such procedure ✌
That 2nd step isn't a good idea.
Do this instead:
x³ - 2*x² + 4*x - 8 = 0
This is a cubic equation with real coefficients, which means there are 3 roots with 1 or 3 real roots. All 3 potential answers show 1 real root and 2 complex conjugate roots. It's easy to test the real solutions of the first 2 answers listed, x = 1 or 2, so let's do that:
x = 1:
1 - 2 + 4 - 8 =?= 0
-5 ≠ 0
x = 2:
8 - 8 + 8 - 8 =?= 0
0 = 0
Thus, we know x = 2 is one root, and the answer must be "b". As a check, we can find the remaining 2 roots by factoring out (x-2) from the equation:
x²*(x-2) + 4*(x-2) = 0
(x²+4)*(x-2) = 0
Now, we know (x²+4) = 0, which means x = ±2*i .
This matches the other values in "b".
Of course, if you had noticed that the coefficients are increasing powers of 2 with alternating signs, that would immediately lead you to factor the equation after rewriting it as:
2⁰*x³ - 2¹*x² + 2²*x¹ - 2³*x⁰ = 0
I wouldn't even know where to start with a cubic involving imaginary numbers. I'm just chuffed I even know what any of that is and that's as far as it goes.
Edit- All I remember is complex numbers are used for oscillations in quantum mechanics because i=i, i2=-1, i3=-i, i4=1,i5=i, etc
Can't I solve by factorization?
Yes you can:
x³ - 2*x² + 4*x - 8 = 0
x²*(x-2) + 4*(x-2) = 0
(x²+4)*(x-2) = 0
Now, we know:
(x²+4) = 0 and (x-2) = 0
x = ±2*i and x = 2
These values match the values in "b".
Of course, if you had noticed that the coefficients are increasing powers of 2 with alternating signs, that would immediately lead you to factor the equation after rewriting it as:
2⁰*x³ - 2¹*x² + 2²*x¹ - 2³*x⁰ = 0
My hair are, indeed, on fire.
But the title is right and my algebra knowledge is not yet that deep.
But I learned a lot of new stuff with this demonstration.
uncle uncle I tap I tap. Lol.
3rd degree equation with(i)
My answer is (b)
Not a terribly difficult problem to pick out the correct roots. Just solve using the real number and them use the imaginary roots if you like a little more difficult challenge. B is the answer.
No bone today.
2q+2q=
I looked at it for 20 seconds and substituted 2 in for x and that was a solution, so i chose b
x^3-2x^2+4x-8=0
x^2(x-2)+4(x-2)=0
(x^2+4)(x-2)=0
I'm gonna cheat...
the two solutions are
x^2=-4 and x=2
x^2 yields 2 solutions for a total of 3.
only option b) has x=2 as solution...
in like manner....
just plug/chug with first 1 and then 5 isn't that involved.
b)
b no math required
B
😂
Teaching negativity from the : Many people will get Wrong.