SOLUTION: Real or imaginary solutions x^3=216 6x=9+x^2 X^2+3x+3=0 X^3+125=0 X^4-24x^2+100=0

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Real or imaginary solutions x^3=216 6x=9+x^2 X^2+3x+3=0 X^3+125=0 X^4-24x^2+100=0      Log On


   



Question 840889: Real or imaginary solutions
x^3=216
6x=9+x^2
X^2+3x+3=0
X^3+125=0
X^4-24x^2+100=0

Found 2 solutions by Fombitz, KMST:
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E3=216 Both
6x=9%2Bx%5E2 Real
x%5E2%2B3x%2B3=0 Imaginary
x%5E3%2B125=0 Both
x%5E4-24x%5E2%2B100=0 Real

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E3=216-->x%5E3-216=0-->x%5E3-6%5E3=0-->%28x-6%29%28x%5E2%2B6x%2B6%5E2%29=0-->%28x-6%29%28x%5E2%2B6x%2B36%29=0
x=6 , which makes %28x-6%29=0 is the real solution;
x%5E2%2B6x%2B36=0 , which has no real solution yields two more imaginary ones.

6x=9%2Bx%5E2-->x%5E2-6x%2B9=0-->%28x-3%29%5E2=0 has x=3 as a double real solution (no imaginary ones).

x%5E2%2B3x%2B3=0 has no real solutions, just imaginary ones.
You would know that by calculating the discriminant.
A quadratic equation, that can be written as
ax%5E2%2Bbx%2Bc=0 , has solutions that can be calculated using the quadratic formula:
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
The expression b%5E2-4%2Aa%2Ac is called the discriminant, and its sign will tell you about the solutions. A negative discriminant means no real solutions. A positive one means two (different) real solutions. If the discriminant is zero, there is just one (double) real solution (as happened for x%5E2-6x%2B9=0 above).
In this case a=1 , b=3 , and c=3 , and
b%5E2-4%2Aa%2Ac=3%5E2%2A4%2A1%2A3=9-12=-3%3C0 , so no real solutions.

x%5E3%2B125=0-->x%5E3%2B5%5E3=0-->%28x%2B5%29%28x%5E2-5x%2B5%5E2%29=0
x=-5 is the real solution; x%5E2-5x%2B25=0 , which has no real solution yields two more imaginary ones.

x%5E4-24x%5E2%2B100=0 is more complicated.
You could use the change of variable y=x%5E2 to write the equation as
y%5E2-24y%2B100=0 , and find that the solutions for y .
You could solve for y by completing the square, or by using the quadratic formula.
Either way, you will find
y=12+%2B-+sqrt%2844%29 (or an equivalent expression).
Since both y solutions are positive, you will have four real solutions to
x%5E2=12+%2B-+sqrt%2844%29 , and those are the solutions of
x%5E4-24x%5E2%2B100=0 .
so there are two real solutions.