Question

What are the solutions of the equation x4 + 3x2 + 2 = 0? Use u substitution to solve. x = plus-or-minus i StartRoot 2 EndRoot and x = ±1 x = plus-or-minus i StartRoot 2 EndRoot and x = ±i x = plus-or-minus StartRoot 2 EndRoot and x = ±i x = plus-or-minus StartRoot 2 EndRoot and x = ±1

Answer 1

Answer:

x = ± i & ± i√2

Step-by-step explanation:

The given equation is x⁴ + 3x² + 2 = 0

let u = x²

∴ u² + 3u + 2 = 0

∴ ( u + 2 ) ( u + 1 ) = 0

∴ u = -1 or -2

∴ x² = -1   or  x² = -2

∴ x = ±√(-1) = ± i

And x = ±√(-2) = ± i √2

Note i = √(-1)

Answer 2

Answer:

$x=\pm\sqrt{2}i\text{ (or) }x=\pm i$

Step-by-step explanation:

We have been given an equation $x^4 + 3x^2 + 2 = 0$. We are asked to find the solutions of our given equation using u-substitution.

We can rewrite our given equation as:  

$(x^2)^2+3x^2+2 = 0$

Let us assume that $u=x^2$.

$u^2+3u+2 = 0$

$u^2+2u+u+2 = 0$

$u(u+2)+1(u+2) = 0$

$(u+2)(u+1) = 0$

$(u+2)=0\text{ (or) }(u+1) = 0$

$u=-2\text{ (or) }u=-1$

Upon substituting back the value of u, we will get:

$x^2=-2\text{ (or) }x^2=-1$

$x=\pm\sqrt{-2}\text{ (or) }x=\pm\sqrt{-1}$

$x=\pm\sqrt{-1\cdot 2}\text{ (or) }x=\pm\sqrt{-1}$

Now we will use imaginary unit i. We know that $i^2=-1$.

$x=\pm\sqrt{i^2\cdot 2}\text{ (or) }x=\pm\sqrt{i^2}$

$x=\pm\sqrt{2}i\text{ (or) }x=\pm i$

Therefore, the solutions for our given equation are $x=\pm\sqrt{2}i\text{ (or) }x=\pm i$.