Question

What are the real and complex solutions of the polynomial equation? 0=x^4+3x^2-4
A. -1,1
B. 1,2i
C. -1,1,-2,2
D. -1,1,-2i, 2i

Answer 1

Answer:

D. -1,1,-2i, 2i

Step-by-step explanation:

We should have 4 roots since this is a 4th degree polynomial.  We just have to watch since some of them may be multiple roots.

x^4+3x^2-4 = 0

Factoring this equation

(x^2 - 1) (x^2+4) = 0

We can factor the first term

(x-1) (x+1) (x^2+4) =0

Using the zero product property

x-1 =0  x+1 = 0  x^2+4 =0

x=1        x=-1        x^2 = -4

                            Taking the square root of each side

                           sqrt(x^2) =± sqrt(-4)

                                     x =± sqrt(-1) sqrt(4)

                                       x =±i *2

                                       x = ±2i

Answer 2

Steps:

So firstly, I will be factoring by grouping. Firstly, what two terms have a product of -4x⁴ and a sum of 3x²? That would be -x² and 4x². Replace 3x² with -x² + 4x²:

$0=x^4-x^2+4x^2-4$

Next, factor x⁴ - x² and 4x² - 4 separately. Make sure that they have the same quantity on the inside:

$0=x^2(x^2-1)+4(x^2-1)$

Now you can rewrite it as:

$0=(x^2+4)(x^2-1)$

However, we can simplify it even further. With the second factor, apply the difference of squares rule (x² - y² = (x + y)(x - y)):

$x^2-1=(x+1)(x-1)\\\\0=(x^2+4)(x+1)(x-1)$

Now it's fully factored. With this, apply the zero product property and solve:

$x^2+4=0\\x^2=-4\\x=\pm \sqrt{-4}=\pm \sqrt{4}i\\x=\pm\ 2i\\x=2i,-2i\\\\x+1=0\\x=-1\\\\x-1=0\\x=1$

Answer

In short, your answer is D. -1, 1, -2i, 2i