I solved it by squaring the first equation. This gives x^4+1/x^4=0. The second equation can be factored in (x^20+1/x^20)(x^20-1/x^20). The first of these factors can be decomposed into (x^4+1/x^4) times another factor starting with x^16. The whole expression is thus a product of three factors. Since x^4+1/x^4 is 0, the whole product is 0 as well
How do we know that x^20 + 1/x^20 can be factored into (x^4 + 1/x^4) * (x^16 +... more terms) without leaving the remainder? Do we have to use remainder theorem to prove that y + 1/y is always a factor of y^n + 1/y^n or something like that, or is there another known theorem out there related to this?
@gametimewitharyan6665 x^20+1/x^20 can be seen as the sum of two uneven powers (x^4)^5+(1/x^4)^5. Sums of uneven powers a^n+b^n can be always factored as the product of two terms: the sum of the bases (in this case a+b) times another factor obtained as the sum of all terms a^(n-1-k)b^k with 0
@@andreabaldacci1142 oh, yes I have seen this result which is proved using remainder theorem Proof - p(x) = x^n + a^n Then p(-a) = (-a)^n - a^n p(a) is 0 when n is an odd integer, and using the remainder theorem if p(-a) = 0 then x+a is a factor of p(x)... Thanks for clarifying!
If I square the original expression it becomes x^4+1/x^+2=2 which simplifies to x^8=-1 Also in the first method once you evaluated x^2 without evaluating x you could have raised it to 20th power to get the 40th power. Just save a couple of steps.
Multiply by x^40 + 1/x^40 to get x^80 + 1/x^80. Calculate this last quantity by repeated multiplication (takes about 5 steps to build up to 80) and we get zero. Which means x^40-1/x^40 = 0.
Thank you for asking! Using the vertex and the y-intercept... let f(x) = ax² + bx + c x-vertex = h = -b/2a y-vertex = f(h) = k (h, k) is the vertex (0, c) is the y- intercept You can also reflect the y-intercept along the axis of symmetry which is the vertical line that goes through the vertex (given by the equation x = -b/2a) to get an additional point. This gives you three points on the parabola. Connect the dots with a smooth curve and that's it...
Just square the given equation to get x⁴ + 2 + 1/x⁴ = 2 ie x⁸ = -1. Then x^40 - 1/x^40 = (x⁸)⁵ - 1/(x⁸)⁵ = -1 - (-1) = 0. One can do all this on one's head. What a waste of time.
Never a waste of time at SyberMath when one takes the time to learn something new.
Thank you! 😍❤️
Squaring both sides of x²+1/x²=√2 leads to x⁴+1/x⁴=0. So x⁴=(-1/x⁴). Raising to power 10 both sides leads to x⁴⁰=1/x⁴⁰
So x⁴⁰-1/x⁴⁰=0
I solved it by squaring the first equation. This gives x^4+1/x^4=0. The second equation can be factored in (x^20+1/x^20)(x^20-1/x^20). The first of these factors can be decomposed into (x^4+1/x^4) times another factor starting with x^16. The whole expression is thus a product of three factors. Since x^4+1/x^4 is 0, the whole product is 0 as well
How do we know that x^20 + 1/x^20 can be factored into (x^4 + 1/x^4) * (x^16 +... more terms) without leaving the remainder?
Do we have to use remainder theorem to prove that y + 1/y is always a factor of y^n + 1/y^n or something like that, or is there another known theorem out there related to this?
@gametimewitharyan6665 x^20+1/x^20 can be seen as the sum of two uneven powers (x^4)^5+(1/x^4)^5. Sums of uneven powers a^n+b^n can be always factored as the product of two terms: the sum of the bases (in this case a+b) times another factor obtained as the sum of all terms a^(n-1-k)b^k with 0
@@andreabaldacci1142 oh, yes I have seen this result which is proved using remainder theorem
Proof -
p(x) = x^n + a^n
Then p(-a) = (-a)^n - a^n
p(a) is 0 when n is an odd integer, and using the remainder theorem
if p(-a) = 0 then x+a is a factor of p(x)...
Thanks for clarifying!
Good thinking
Loved the second method!
Glad you liked it!
If I square the original expression it becomes x^4+1/x^+2=2 which simplifies to x^8=-1
Also in the first method once you evaluated x^2 without evaluating x you could have raised it to 20th power to get the 40th power.
Just save a couple of steps.
nice one!
Thanks
Multiply by x^40 + 1/x^40 to get x^80 + 1/x^80.
Calculate this last quantity by repeated multiplication (takes about 5 steps to build up to 80) and we get zero. Which means x^40-1/x^40 = 0.
Thank you Master...😭
Np
👍
Excuse me, can you tell me how to draw quadratic functions that have complex roots
Thank you for asking!
Using the vertex and the y-intercept...
let f(x) = ax² + bx + c
x-vertex = h = -b/2a
y-vertex = f(h) = k
(h, k) is the vertex
(0, c) is the y- intercept
You can also reflect the y-intercept along the axis of symmetry which is the vertical line that goes through the vertex (given by the equation x = -b/2a) to get an additional point.
This gives you three points on the parabola. Connect the dots with a smooth curve and that's it...
@@SyberMath
Thanks for answering...
Integrate (tan²x + sec²×)/(csc²×+cot²x)
I also got zero.
Nice
Just square the given equation to get x⁴ + 2 + 1/x⁴ = 2 ie x⁸ = -1. Then x^40 - 1/x^40 = (x⁸)⁵ - 1/(x⁸)⁵ = -1 - (-1) = 0. One can do all this on one's head. What a waste of time.
You have found a good solution so 👍 for that but thus channel is all about different methods. Never a waste of time.
👏