The equation x⁴ + 12x² + 32 = 0 has no real root.
I suppose it is x⁴ - 12x² + 32 = 0
x⁴ - 12x² + 32 = 0
(x²)² - 12x² + 32 = 0
[x² - 4] [x² - 8] = 0
[x² - 2²] [x² - (2√2)²] = 0
(x - 2)(x + 2)(x - 2√2)(x + 2√2) = 0
x = 2, -2, 2√2, -2√2
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For the equation x⁴ + 12x² + 32 = 0,
the four roots are x = 2i, -2i, (2√2)i, -(2√2)i
where i = √(-1)
The highest power of x is 4. Therefore it will have 4 roots.
A basic theorem in algebra is that a function has as many roots as the highest power in the expression of the function.
These roots can be real or complex, and, sometimes, roots are duplicated.
So, f(x) = ax + b has one root
f(x) = ax^2 + bx + c has two roots; these are either both real, both complex or a "double root" -- the same value for both roots.
If the highest power is odd, there will always be at least one real root, If there is one complex root, then there will always be a conjugate complex root as well.
Now, your problem:
x^4 - 12x^2 + 32
To start off, lets sove for x^2 (since x^4 = (x^2)^2)
Let y = x^2
==> y^2 - 12y + 32;
roots are when the expression = 0
so
y = (12 +/- sqrt(144 - 128))/2 = 6 +/- 2 = 4 or 8
so x^2 = 4 or 8
hence x^2 = 4 ==> x = +/- 2 or x = +/- sqrt(8) = +/- 2sqrt(2)
and you have all four of the roots as expected.
Your English grammar sucks!
The root of a function is a value that makes the function zero. You can simply plug in the negative answers to see that they do indeed make that polynomial equal zero.
The question is, what method did you use that didn't produce four solutions?
If you factored it, you should get g(x) = (x-2)(x+2)(x-2√2)(x+2√2)
If you applied the quadratic formula, you should get x^2=4 or x^2=8.
If you graphed the function you would see it cross the x-axis four times.
The Fundamental Theorem of Algebra requires an equation of 4th degree MUST have 4 roots (although they might be complex as well)
x^4-12x^2+32=0 should have 4 roots because
the degree of it is 4. Actually,
x^4-12x^2+32=0
=>
(x^2-4)(x^2-8)=0
=>
x^2=4
x=+/-2
or
x^2=8
x=+/-2sqr(2)
totally 4 real roots.
comment : every nth degree polynomial has n roots...here ±2i & ± √8 i
x⁴ and x² will produce the same results with a positive or a negative value for x. So any root would actually be two, + and –
x⁴ + 12x² + 32 = 0
let z = x²
z² + 12z + 32 = 0
z = –4, –8
if z = –4
x = √–4 = ±2i
if z = –8
x = √8 = ±2i√2
I have no idea how you got your answer, all the roots are imaginary.
I suspect it is
–x⁴ + 12x² – 32 = 0
or
x⁴ – 12x² + 32 = 0
in which case the solutions are
z = 4,8
x = ±2, ±2√2
check each of the 4 in the original equation, you will see that they satisfy it.
I'll do –2
–x⁴ + 12x² – 32 = 0
–(–2)⁴ + 12(–2)² – 32 = 0
–16 + 48 – 32 = 0
ok
If a root works, then root^2 is the same value as (-root)^2. And root^4 is the same value as (-root)^4.
It is a fourth order expression.
x^4-12x^?+32
x⁴ - 12x² + 32 = 0
x⁴ - 12x² = - 32
x⁴ - 12x² + 36 = - 32 + 36
x⁴ - 12x² + 36 = 4
(x² - 6)² = 4
x² - 6 = ± 2
x² = 6 ± 2
First case: x² = 4
x = ± 2
Second case: x² = 8
x = ± 2√2
you are stuck in a two dimension univrrsie.
g(x) = x^4 + 12 x^2 + 32
g(x) = (x^2 + 4) (x^2 + 8)
g(x) = (2 sqrt(2) - i x) (2 sqrt(2) + i x) (x - 2 i) (x + 2 i)
