Respuesta :
This would be easiest to do by completing the square.
x^2+10x-2=0
x^2+10x=2
x^2+10x+25=27
(x+5)^2=27
x+5=±√27
x=-5±√27
(x+5+√27)(x+5-√27)
x^2+10x-2=0
x^2+10x=2
x^2+10x+25=27
(x+5)^2=27
x+5=±√27
x=-5±√27
(x+5+√27)(x+5-√27)
You would use the quadric formula, which you can find on the site, or on google :)
like this:
x= (-10 +/- sqrt(10^2 - 4*1*-2))/2*1
x = -10 +/- sqrt(108)/2
x = (-10 +/- 10.4)/2
x_1 = 0.4/2 = 0.2
x_2 = -20.4/2 = -10.2
The result can be written in the following way, to be a factorized version of the first equation.
(x-x_1)(x-x_2) = 0
Which leads us to the answer:
(x-0.4)(x+10.2) = 0
Youre welcome.
(The answer is not exact, but it is more than close enough.)
like this:
x= (-10 +/- sqrt(10^2 - 4*1*-2))/2*1
x = -10 +/- sqrt(108)/2
x = (-10 +/- 10.4)/2
x_1 = 0.4/2 = 0.2
x_2 = -20.4/2 = -10.2
The result can be written in the following way, to be a factorized version of the first equation.
(x-x_1)(x-x_2) = 0
Which leads us to the answer:
(x-0.4)(x+10.2) = 0
Youre welcome.
(The answer is not exact, but it is more than close enough.)