2007-09-06 09:25:00 · 5 answers · asked by Keith S 1 in Science & Mathematics ➔ Mathematics
You can't using rational factors.
2007-09-06 09:36:19 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
Solve 5x^2+2x-1 = 0 using the quadratic formula.
x = ( - 2 +/- sqrt(4 + 20) ) / (2 * 5)
x = ( - 2 +/- sqrt(4)sqrt(6) ) / 10
x = ( - 2 +/- 2sqrt(6) ) / 10
x = ( - 1 +/- sqrt(6) ) / 5
The factors of 5x^+2x-1 are then:
( x + ( 1 + sqrt(6) ) / 5 )(x + (1 - sqrt(6) ) / 5 )
(1/5)(5x + 1 + sqrt(6))(5x + 1 - sqrt(6)).
2007-09-06 16:37:13 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Use the quadratic formula as follows: -
a= 5, b= 2, c= -1
x = [-b ± â (b^2 - 4ac)] / 2a
x = {-2 ± â [2^2 - 4(5)(-1)] } / 2(5)
x = [-2 ± â (4 + 20)] / 10
x = (-2 ± â24 ) / 10
x = (-2 + â24)/10 or x = (-2 - â24)/10
Using the calculator: -
x = 0.289897948 or x = -0.689897948
2007-09-06 16:49:13 · answer #3 · answered by plolol 2 · 0⤊ 0⤋
I guess you mean 5x^2+2x-1 = 5(x^2 + (2/5)x - 1/5)= 5((x + 1/5)^2 -1/25 -1/5) = 5((x + 1/5)^2 - 6/25) = 5((x+1/5)^2 - (sqrt(6/25))^2) = 5(x + 1/5 + sqrt(6/25))(x + 1/5 - sqrt(6/25)) = 5(x + (1+sqrt(6))/5)(x + (1-sqrt(6))/5)
2007-09-06 16:34:23 · answer #4 · answered by Christophe G 4 · 0⤊ 0⤋
Use the quadratic equation.
2007-09-06 16:31:19 · answer #5 · answered by Shanna J 4 · 0⤊ 0⤋