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nignag [31]
3 years ago
5

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

d 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
Mathematics
2 answers:
Rom4ik [11]3 years ago
5 0

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.

igor_vitrenko [27]3 years ago
5 0

Answer:

<u>x = ± i & ± i√2</u>

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)

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