Question

Show that x^7-3x^4+2x^3-1 = 0 has at least four imaginary roots .​

Answer 1

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Answer 2

Answer:

Give equation,

${x}^{7} - 3 {x}^{4} + 2 {x}^{3} - 1 = 0$

Let,

$f(x) = {x}^{7} - 3 {x}^{4} + 2 {x}^{3} - 1$

f(x) has three changes of signs. Hence there can be a maximum of 3 positive roots.

$f( - x) = {( - x)}^{7} - 3 {( - x)}^{4} + 2 {( - x)}^{3} - 1$

So,

$f( - x) = - {x}^{7} - 3 {x}^{4} - 3 {x}^{3 } - 1$

which has no changes of signs. Hence the given equation has zero negative roots.

Now, as the equation is of 7 th degree, it must have at least (7−3)=4 imaginary roots.

Hence, given equation has at least four imaginary roots (proved)

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549