Solve x2 + 12x + 6 = 0 using the completing-the-square method.

1 answer:

7 0

X^2+12x+6=0 subtract 6 from both sides...

x^2+12x=-6 halve linear coefficient, square it, and add it to both sides...

x^2+12x+36=30 now the left side is a perfect square...

(x+6)^2=30 take square root of both sides...

x+6=30^(1/2) subtract 6 from both sides...

x=-6-30^(1/2) and -6+30^(1/2)

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Angelina_Jolie [31]

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A farmer wants to fence in a rectangular plot of land adjacent to the north wall of his barn. No fencing is needed along the bar

Luba_88 [7]

The dimensions for the plot that would enclose the most area are a length and a width of 125 feet.

In this question we shall use the first and second derivative tests to determine the optimal dimensions of a rectangular plot of land. The perimeter ( $p$ ), in feet, and the area of the rectangular plot ( $A$ ), in square feet, of land are described below:

$p = 2\cdot (w+l)$ (1)

$A = w\cdot l$ (2)

Where:

$w$ - Width, in feet.
$l$ - Length, in feet.

In addition, the cost of fencing of the rectangular plot ( $C$ ), in monetary units, is:

$C = c\cdot p$ (3)

Where $c$ is the fencing unit cost, in monetary units per foot.

Now we apply (2) and (3) in (1):

$p = 2\cdot \left(\frac{A}{l}+l \right)$

$\frac{C}{c} = 2\cdot (\frac{A}{l}+l )$

$\frac{C\cdot l}{c} = 2\cdot (A+l^{2})$

$\frac{C\cdot l}{c}-2\cdot l^{2} = 2\cdot A$

$\frac{C\cdot l}{2\cdot c} - l^{2} = A$ (4)

We notice that fencing costs are directly proportional to the area to be fenced. Let suppose that cost is the maximum allowable and we proceed to perform the first and second derivative tests:

FDT

$\frac{C}{2\cdot c}-2\cdot l = 0$