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IRISSAK [1]
3 years ago
10

How to get x when x^3-x^2+x-1=0?

Mathematics
1 answer:
Art [367]3 years ago
6 0

Answer:

x = 1 or x = ± i

Step-by-step explanation:

Note the sum of the coefficients

1 - 1 + 1 - 1 = 0

This indicates that x = 1 is a root, thus (x - 1) is a factor

Using long division or synthetic division, then

x³ - x² + x - 1 = (x - 1)(x² + 1), thus

(x - 1)(x² + 1) = 0

Equate each factor to zero and solve for x

x - 1 = 0 ⇒ x = 1

x² + 1 = 0 ⇒ x² = - 1 ⇒ x = ± \sqrt{-1} = ± i

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Alexxandr [17]

Given:

The equation is

x^2+(2\sqrt{5})+2x=-10

To find:

The number of roots and discriminant of the given equation.

Solution:

We have,

x^2+(2\sqrt{5})x+2x=-10

The highest degree of given equation is 2. So, the number of roots is also 2.

It can be written as

x^2+(2\sqrt{5}+2)x+10=0

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Discriminant of the given equation is

D=b^2-4ac

D=(2\sqrt{5}+2)^2-4(1)(10)

D=20+8\sqrt{5}+4-40

D=8\sqrt{5}-16

D\approx 1.89>0

Since discriminant is 8\sqrt{5}-16\approx 1.89, which is greater than 0, therefore, the given equation has two distinct real roots.

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Answer:

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