I did something different I used x⁴+2kx²+k²=2kx²-7x+12+k² x⁴-2kx²+k²=-2kx²-7x+12-k² k=0.5 Hence I root squared the equation on both sides getting x²+7.5-12=0 x²-6.5-12=0 Hence I used the quadratic formula to find four solutions
It's worth mentioning that the other roots of k^3+48k-49 will lead to the same four roots for x. We know this, even without working each case out, from the fundamental theorem of algebra.
At 7:55 the cubic polynomial would also have two complex solutions, and therefore four complex a values. Does this mean that the original quartic can also be factored with complex coefficients? How would that look like? Ultimately the roots would need to be the same.
Well we do know all four roots so we write the polynomial as (1/16) (2x + 1 + √13) (2x + 1-- √13) (2x - 1 + i√15) (2x - 1 - i√15) = 0 If you like you can multiply real and complex roots to form two quadratic factors respectively - each with complex coefficients.
I did something different
I used x⁴+2kx²+k²=2kx²-7x+12+k²
x⁴-2kx²+k²=-2kx²-7x+12-k²
k=0.5
Hence I root squared the equation on both sides getting x²+7.5-12=0
x²-6.5-12=0
Hence I used the quadratic formula to find four solutions
It's worth mentioning that the other roots of k^3+48k-49 will lead to the same four roots for x. We know this, even without working each case out, from the fundamental theorem of algebra.
yes -- these being complex roots
Love the solution!
I'm in standard 10th but i understand this clearly
At 7:55 the cubic polynomial would also have two complex solutions, and therefore four complex a values.
Does this mean that the original quartic can also be factored with complex coefficients? How would that look like?
Ultimately the roots would need to be the same.
Well we do know all four roots so we write the polynomial as (1/16) (2x + 1 + √13) (2x + 1-- √13) (2x - 1 + i√15) (2x - 1 - i√15) = 0
If you like you can multiply real and complex roots to form two quadratic factors respectively - each with complex coefficients.