SOLUTION: Hello there, I'm trying to arrive at the correct answer of {±4,±i√3}. Any help would be great thank you. Solve and express your solution in simplified form. x^4−13x

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Hello there, I'm trying to arrive at the correct answer of {±4,±i√3}. Any help would be great thank you. Solve and express your solution in simplified form. x^4−13x      Log On


   

Question 1178897: Hello there,
I'm trying to arrive at the correct answer of {±4,±i√3}. Any help would be great thank you.

Solve and express your solution in simplified form.
x^4−13x^2−48=0
Found 3 solutions by Boreal, MathLover1, Solver92311:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
Treat this like a quadratic, since the first term is a power of 4, the middle term a power of 2 and the last term a constant.
(x^2-16)(x^2+3)=0
x=+/-4 and +/- i sqrt(3)

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

x%5E4-13x%5E2-48=0...factor
x%5E4-16x%5E2%2B3x%5E2-48=0+,,,group
%28x%5E4-16x%5E2%29%2B%283x%5E2-48%29=0+
x%5E2%28x%5E2-16%29%2B3%28x%5E2-16%29=0+
%28x%5E2+-16%29%28x%5E2+%2B+3%29+=+0
%28x%5E2+-4%5E2%29%28x%5E2+%2B+3%29+=+0
%28x+-+4%29+%28x+%2B+4%29+%28x%5E2+%2B+3%29+=+0

solutions:
if %28x+-+4%29++=+0=>x=4
if +%28x+%2B+4%29++=+0=> x=-4
if +%28x%5E2+%2B+3%29+=+0=>+x%5E2+=-+3=>+x+=sqrt%28-+3%29 =>+x+=i%2Asqrt%283%29 or +x+=+-i%2Asqrt%283%29
so, we have:
Real solutions:
x=4
x=-4
Complex solutions:
+x+=i%2Asqrt%283%29
+x+=+-i%2Asqrt%283%29


Answer by Solver92311(821) About Me  (Show Source):
You can put this solution on YOUR website!


Let , then is equivalent to

Use the quadratic formula to solve for the two values of . Then solve for each value of giving you four distinct roots as is expected from a quartic equation. Hint:

John

My calculator said it, I believe it, that settles it

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