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3 years ago

Solve x2 + 9x + 8 = 0 by completing the square. What are the solutions?

1 answer:

8 0

Answer: (-1,-8)

Explanation:

x^2 + 9x + 8 = 0
x^2 + 9x = -8

Add (9/2)^2 to both sides

x^2 + 9x + (9/2)^2 = -8 + (9/2)^2
(x + 9/2)^2 = -8 + 81/4
(x + 9/2)^2 = -32/4 + 81/4
(x + 9/2)^2 = 49/4

Square root both sides

Sqrt (x + 9/2)^2 = sqrt 49/4
x + 9/2 = plus or minus 7/2

Set x + 9/2 = 7/2:

x + 9/2 = 7/2
x = 7/2 - 9/2
x = -2/2 = -1

Set x + 9/2 = -7/2:
x + 9/2 = -7/2
x = -7/2 - 9/2
x = -16/2 = -8

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3. You have a square plot of land that has an area of 144 square feet. You

ElenaW [278]

Answer:

113.04square feet

Step-by-step explanation:

square root of 144 is 12

use it as the dimensions of the plot.

the flower gardens largest possible diameter is 12

so divide 12 by a half

6=radius

3.14 times 6squared

you get 113.04

3 0

4 years ago

The perimeter of a rectangle is 70 cm. The ratio of length to width is 2:5. Find the length and width of the rectangle.

olya-2409 [2.1K]

Answer:

length = 10 cm; width = 25 cm

Step-by-step explanation:

Let's call the length 2x and the width 5x. Since perimeter can be calculated by multiplying the sum of the length and width by 2 we can write:

2 * (2x + 5x) = 70

2 * (7x) = 70

7x = 35

x = 5 which means the length is 2 * 5 = 10 and the width is 5 * 5 = 25.

5 0

3 years ago

The sides of a triangle are 1, x, and x2. what are possible values of x?

Whitepunk [10]

The sides of the triangle are given as 1, x, and x².

The principle of triangle inequality requires that the sum of the lengths of any two sides should be equal to, or greater than the third side.

Consider 3 cases
Case (a): x < 1,
Then in decreasing size, the lengths are 1, x, and x².
We require that x² + x ≥ 1
Solve x² + x - 1 =
x = 0.5[-1 +/- √(1+4)] = 0.618 or -1.618.
Reject the negative length.
Therefore, the lengths are 0.382, 0.618 and 1.

Case (b): x = 1
This creates an equilateral triangle with equal sides
The sides are 1, 1 and 1.

Case (c): x>1
In increasing order, the lengths are 1, x, and x².
We require that x + 1 ≥ x²
Solve x² - x - 1 = 0
x = 0.5[1 +/- √(1+4)] = 1.6118 or -0.618
Reject the negative answr.
The lengths are 1, 1.618 and 2.618.

Answer:
The possible lengths of the sides are
(a) 0.382, 0.618 and 1
(b) 1, 1 and 1.
(c) 2.618, 1.618 and 1.

7 0

4 years ago

*Asymptotes*<br> g(x) =2x+1/x-3 <br><br> Give the domain and x and y intercepts

Nataly [62]

Answer: Assuming the function is $g(x)=\frac{2x+1}{x-3}$ :

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The horizontal asymptote is $y=2$ .

The vertical asymptote is $x=3$ .

Step-by-step explanation:

I'm going to assume the function is: $g(x)=\frac{2x+1}{x-3}$ and not $g(x)=2x+\frac{1}{x}-3$ .

So we are looking at $g(x)=\frac{2x+1}{x-3}$ .

The x-intercept is when y is 0 (when g(x) is 0).

Replace g(x) with 0.

$0=\frac{2x+1}{x-3}$

A fraction is only 0 when it's numerator is 0. You are really just solving:

$0=2x+1$

Subtract 1 on both sides:

$-1=2x$

Divide both sides by 2:

$\frac{-1}{2}=x$

The x-intercept is $(\frac{-1}{2},0)$ .

The y-intercept is when x is 0.

Replace x with 0.

$g(0)=\frac{2(0)+1}{0-3}$

$y=\frac{2(0)+1}{0-3}$

$y=\frac{0+1}{-3}$

$y=\frac{1}{-3}$

$y=-\frac{1}{3}$ .

The y-intercept is $(0,\frac{-1}{3})$ .

The vertical asymptote is when the denominator is 0 without making the top 0 also.

So the deliminator is 0 when x-3=0.

Solve x-3=0.

Add 3 on both sides:

x=3

Plugging 3 into the top gives 2(3)+1=6+1=7.

So we have a vertical asymptote at x=3.

Now let's look at the horizontal asymptote.

I could tell you if the degrees match that the horizontal asymptote is just the leading coefficient of the top over the leading coefficient of the bottom which means are horizontal asymptote is $y=\frac{2}{1}$ . After simplifying you could just say the horizontal asymptote is $y=2$ .

Or!

I could do some division to make it more clear. The way I'm going to do this certain division is rewriting the top in terms of (x-3).

$y=\frac{2x+1}{x-3}=\frac{2(x-3)+7}{x-3}=\frac{2(x-3)}{x-3}+\frac{7}{x-3}$

$y=2+\frac{7}{x-3}$

So you can think it like this what value will y never be here.

7/(x-3) will never be 0 because 7 will never be 0.

So y will never be 2+0=2.

The horizontal asymptote is y=2.

(Disclaimer: There are some functions that will cross over their horizontal asymptote early on.)

6 0

3 years ago

Find the product 3 • ( -7 )

Vera_Pavlovna [14]

Answer:-21

Step-by-step explanation:

3•(-7)

3 x -7=-21

4 0

3 years ago