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goldenfox [79]
3 years ago
8

Find the number of roots for each equation. 5. 5x4 + 12x3 – x2 + 3x + 5 = 0

Mathematics
1 answer:
pav-90 [236]3 years ago
7 0

Answer:

The number of roots for  equation 5x^4 + 12x^3 – x^2 + 3x + 5 = 0 is 4 .

Step-by-step explanation:

Here, the given function polynomial is :

P(x) : 5x^4 + 12x^3 – x^2 + 3x + 5 = 0

The Fundamental Theorem of Algebra says that a polynomial of degree n will have exactly n roots (counting multiplicity).

Now here, the degree if the polynomial is 4 (highest power of variable x).

So, according to the Fundamental Theorem, the given polynomial can have AT MOST 4 roots, counting Multiplicity.

Hence,  the number of roots for  equation 5x^4 + 12x^3 – x^2 + 3x + 5 = 0 is 4 .

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3 years ago
Find x please show work if u can​
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Answer:

x is 12

Step-by-step explanation:

You can find x two different ways.

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Step-by-step explanation:

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Read 2 more answers
A binary operation is defined on the set of real numbers ℝ by
zloy xaker [14]

Answer:

I. m = 2401

II. ((n+1) ∆ y)/n = 1/n[(n – y + 2)(n – y) + 1]

Step-by-step explanation:

I. Determination of m

x ∆ y = x² − 2xy + y²

2 ∆ − 5 = √m

2² − 2(2 × –5) + (–5)² = √m

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Take the square of both side

49² = m

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II. Simplify ((n+1) ∆ y)/n

We'll begin by obtaining (n+1) ∆ y. This can be obtained as follow:

x ∆ y = x² − 2xy + y²

(n+1) ∆ y = (n+1)² – 2(n+1)y + y²

(n+1) ∆ y = n² + 2n + 1 – 2ny – 2y + y²

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(n+1) ∆ y = (n – y + 2)(n – y) + 1

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((n+1) ∆ y)/n = 1/n[(n – y + 2)(n – y) + 1]

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Step-by-step explanation:

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