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3 years ago

For what value of x does 25x=5x2-3?

2 answers:

3 0

I might be wrong but I'll give it a shot, you subtract the 5x from the right side and put it on the left with makes it 20x=2-3. Then subtract the 3 from the 2 which makes it 20x=-1. So it would be -0.05.

I'm not 100% sure. But his is my guess.

Savatey [412]3 years ago

3 0

Important: Use " ^ " to indicate exponentiation: 25x=5x^2-3.

Rearrange these terms in descending powers of x: 5x^2 - 25x - 3 = 0

This is a quadratic equation. To solve it, you could use a graphing calculator, the quadratic formula, completing the square or other approaches.

Let's complete the square of 5x^2 - 25x - 3 = 0:

5(x^2 - 5x) = 3
5(x^2 - 5x + 25/4 - 25/4) = 3
5(x-5/2)^2 - 125/4 = 3 = 12/4
5(x-5/2)^2 =137/4
(x-5/2)^2 -25/4 = 137/20
(x -5/2)^2 = 125/20+137/20 = 162/20= 81/10
Taking the sqrt of both sides, we get:

x-5/2 = plus or minus 9/sqrt(10)
Then:

x = 5/2 plus or minus 9sqrt(10) / 10, or
x = 25/10 plus or minus 9sqrt(10)/10, or

25 plus or minus 9sqrt(10)
x = -------------------------------------
10

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According to the Rational Root Theorem, which function has the same set of potential rational roots as the function g(x) = 3x^5

nata0808 [166]

Answer:

The correct option is A.

Step-by-step explanation:

The question is:

According to the Rational Root Theorem, which function has the same set of potential rational roots as the function g(x) = 3x5 – 2x4 + 9x3 – x2 + 12?

A).f(x) = 3x5 – 2x4 – 9x3 + x2 – 12

B).f(x) = 3x6 – 2x5 + 9x4 – x3 + 12x

C).f(x) = 12x5 – 2x4 + 9x3 – x2 + 3

D).f(x) = 12x5 – 8x4 + 36x3 – 4x2 + 48

Solution:

The function given to us is:

3x5 – 2x4 + 9x3 – x2 + 12

Find the factors of 12 and consider it as 'p'

The factors of 12 are:

p = +/- 1 , +/-2 , +/-3 , +/- 4 , +/-6

Now find the factors of 3 and consider it as 'q'

The factors of 3 are:

q = +/- 1 , +/- 3

We know that we write rational terms in p/q form.

Therefore the Rational roots are given by p/q

+/- 1 , +/-2 , +/-3 , +/- 4 , +/-6 , +/- 1/3 , +/- 2/3 , +/- 4/3

Now we will solve the first function given in part A.

f(x) = 3x^5 – 2x^4 - 9x^3 + x^2 - 12

Again find the factors of 12 and consider it as 'p'

The factors are:

p = +/- 1 , +/-2 , +/-3 , +/- 4 , +/-6

Now find the factors of 3 and consider it as 'q' .

q = +/- 1 , +/- 3

Rational root are given by p/q

+/- 1 , +/-2 , +/-3 , +/- 4 , +/-6 , +/- 1/3 , +/- 2/3 , +/- 4/3

Therefore the rational roots of the given function matches the rational roots of the given equation.

Hence the correct option is A

6 0

2 years ago

What is the degree 3 polynomial that has zeros -3,2 and 5 and in which the coefficient of x^2 is -8?

Setler [38]

For zeroes/roots r1,r2, r3
the facors of the polynomial is
(x-r1)(x-r2)(x-r3)

roots
-3,2,5
(x-(-3))(x-2)(x-5)
(x+3)(x-2)(x-5)
the coefient of the x term is -4
so mutliply the whole thing by 2

(2)(x+3)(x-2)(x-5)

expand
f(x)=2x^3-8x^2-22x+60

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3 years ago

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EleoNora [17]

Answer:

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2 years ago

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Pavel [41]

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3 years ago

Read 2 more answers

The focus of a parabola is (−5,−1) and the directrix is y=−3. What is an equation of the parabola? (x+5)2=8(y+5)

Anna35 [415]

Graph x2 = 4y and state the vertex, focus, axis of symmetry, and directrix.This is the same graphing that I've done in the past: y = (1/4)x2. So I'll do the graph as usual: The vertex is obviously at the origin, but I need to "show" this "algebraically" by rearranging the given equation into the conics form:x2 = 4y Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved
(x – 0)2 = 4(y – 0)This rearrangement "shows" that the vertex is at (h, k) = (0, 0). The axis of symmetry is the vertical line right through the vertex: x = 0. (I can always check my graph, if I'm not sure about this.) The focus is "p" units from the vertex. Since the focus is "inside" the parabola and since this is a "right side up" graph, the focus has to be above the vertex.From the conics form of the equation, shown above, I look at what's multiplied on the unsquaredpart and see that 4p = 4, so p = 1. Then the focus is one unit above the vertex, at (0, 1), and the directrix is the horizontal line y = –1, one unit below the vertex.vertex: (0, 0); focus: (0, 1); axis of symmetry: x = 0; directrix: y = –1Graph y2 + 10y + x + 25 = 0, and state the vertex, focus, axis of symmetry, and directrix.Since the y is squared in this equation, rather than the x, then this is a "sideways" parabola. To graph, I'll do my T-chart backwards, picking y-values first and then finding the corresponding x-values for x = –y2 – 10y – 25:To convert the equation into conics form and find the exact vertex, etc, I'll need to convert the equation to perfect-square form. In this case, the squared side is already a perfect square, so:y2 + 10y + 25 = –x 
(y + 5)2 = –1(x – 0)This tells me that 4p = –1, so p = –1/4. Since the parabola opens to the left, then the focus is 1/4 units to the left of the vertex. I can see from the equation above that the vertex is at (h, k) = (0, –5), so then the focus must be at (–1/4, –5). The parabola is sideways, so the axis of symmetry is, too. The directrix, being perpendicular to the axis of symmetry, is then vertical, and is 1/4 units to the right of the vertex. Putting this all together, I get:vertex: (0, –5); focus: (–1/4, –5); axis of symmetry: y = –5; directrix: x = 1/4Find the vertex and focus of y2 + 6y + 12x – 15 = 0The y part is squared, so this is a sideways parabola. I'll get the y stuff by itself on one side of the equation, and then complete the square to convert this to conics form.y2 + 6y – 15 = –12x 
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(y + 3)2 – 15 = –12x + 9 
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(y + 3)2 = –12(x – 2) 
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8 0

2 years ago