Solve 5x^2 + 2x + 1 = 0?
回答 (5)
2008-09-15 4:08 pm
✔ 最佳答案
5x^2 + 2x + 1 = 0use quadratic formula
x = [-2 ±(√4 - 20)]/10
x = [-2 ±4i]/10
x = (-1 ±2i)/5
2008-09-16 1:27 am
x = [ - 2 ± â (4 - 20 ) ] / 10
x = [ - 2 ± â ( - 16 ) ] / 10
x = [ - 2 ± 4 i ] / 10
x = [ - 1 ± 2 i ] / 5
x = [ - 2 ± â ( - 16 ) ] / 10
x = [ - 2 ± 4 i ] / 10
x = [ - 1 ± 2 i ] / 5
2008-09-15 11:10 pm
This equation is of form ax^2+bx+c
a = 5 b = 2 c = 1
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[-2 +/-sqrt(2^2-4(5)(1)]/(2)(5)
discriminant is b^2-4ac =-16
i^2 = -1, so âi^2 = i
No real roots: The complex roots are
x=[-2 +i â(16)] / (2)(5)
x=[-2 -i â(16)] / (2)(5)
x=[-2+i4] / 10 =-1/5+2/5i
x=[-2-i4] / 10 =-1/5-2/5i
a = 5 b = 2 c = 1
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[-2 +/-sqrt(2^2-4(5)(1)]/(2)(5)
discriminant is b^2-4ac =-16
i^2 = -1, so âi^2 = i
No real roots: The complex roots are
x=[-2 +i â(16)] / (2)(5)
x=[-2 -i â(16)] / (2)(5)
x=[-2+i4] / 10 =-1/5+2/5i
x=[-2-i4] / 10 =-1/5-2/5i
2008-09-15 11:09 pm
5x^2 + 2x + 1 = 0
x = [-b 屉(b^2 - 4ac)]/2a
a = 5
b = 2
c = 1
x = [-2 屉(4 - 20)]/10
x = [-2 屉-16]/10 (imaginary number)
(no real roots)
x = [-b 屉(b^2 - 4ac)]/2a
a = 5
b = 2
c = 1
x = [-2 屉(4 - 20)]/10
x = [-2 屉-16]/10 (imaginary number)
(no real roots)
2008-09-15 11:08 pm
use quadratic formula equation!!! (-b +/- square root(b^2 - 4ac))/(2a)
in this case a = 5, b = 2, c = 1;
(-2 +/- square root(2^2 - 4*5*1)) / (2*5)
(-2 +/- square root(4 - 20)) / (10)
(-2 +/- square root(-16)) / (10) //since is -16 is imagin
(-2 +/- 4i) / (10)
(-0.2 + 0.4i) <<one answer of x
(-0.2 - 0.4i ) <<second answer of x
in this case a = 5, b = 2, c = 1;
(-2 +/- square root(2^2 - 4*5*1)) / (2*5)
(-2 +/- square root(4 - 20)) / (10)
(-2 +/- square root(-16)) / (10) //since is -16 is imagin
(-2 +/- 4i) / (10)
(-0.2 + 0.4i) <<one answer of x
(-0.2 - 0.4i ) <<second answer of x
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