Hi,
I'm trying to help a high-school sophomore with a math problem, and unfortunately my algebra days are long behind me. Here's the equation:
x^3-9x-440=0
I know x=8, but I don't know how to find it. I'd appreciate some guidance.
Thanks.
Hi,
I'm trying to help a high-school sophomore with a math problem, and unfortunately my algebra days are long behind me. Here's the equation:
x^3-9x-440=0
I know x=8, but I don't know how to find it. I'd appreciate some guidance.
Thanks.
use the rational root theorem. if x an integer is a zero it is factor of 440
so look at them 1,2,4,5,8,10,11,20,22,40,44,55,88,110,220,440 ( both plus and minus)
as the sign changes once so there is a positive root
x^3 = 440 + 9x or x^3 > 440 or x > 7
we need to check for values > 7( from the above set)
try 8 and it succeed
so we need not try further
you can devide by x-8 and get a quadratic and solve the same
Knowing that $x=8$, we could them proceed to rewrite the equation as:
$ \displaystyle x^3-8x^2+8x^2-64x+55x-440=0$
Factor:
$ \displaystyle x^2(x-8)+8x(x-8)+55(x-8)=0$
$ \displaystyle (x-8)\left(x^2+8x+55\right)=0$
To get the other 2 roots, we apply the quadratic formula to the quadratic factor:
$ \displaystyle x=\frac{-8\pm\sqrt{8^2-4(1)(55)}}{2(1)}=-4\pm\sqrt{39}i$
$x^3 - 9x -440$
$=x^3 - 8x^2 + 8x^2 - 64x + 55x - 440$
$=x^2(x - 8) + 8x(x - 8) + 55(x - 8)$
$=(x - 8)(x^2 + 8x + 55)$
$=(x - 8)(x^2 + 8x + 16 -16 + 55)$
$=(x - 8)((x + 4)^2 + 39)$
$=(x - 8)(x + 4)^2 - (-39))$
$=(x - 8)(x + 4)^2 - (\sqrt{39}i)^2$
$=(x - 8)(x + 4 - \sqrt{39}i)(x + 4 + \sqrt{39}i)$