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dusya [7]
3 years ago
5

Find the solutions to the equation below 30x^2-28x+6 =0

Mathematics
2 answers:
Nina [5.8K]3 years ago
8 0

Answer:

Option A and E are the correct options.

Step-by-step explanation:

30x²-28x+6=0

2(15x²-14x+3) = 0

Or 15x²-14x+3 = 0

15x²-9x-5x+3 = 0

3x(x-3) - 1(5x-3) = 0

(3x-1)(5x-3) = 0

Now by the zero product rule

3x-1 = 0

3x = 1

x = 1/3

And (5x-3) = 0

5x = 3

x = 3/5

So two solutions are (1/3, 3/5).

Therefore options A and E are the correct options.

Naddik [55]3 years ago
6 0

For this case, we have the following equation of the second degree:

30x ^ 2-28x + 6 = 0

If we divide between 2 on both sides of the equation, we will have:

15x ^ 2-14x + 3 = 0

Where:

a = 15\\b = -14\\c = 3

The solutions will come from:

x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}

Substituting:

x = \frac {- (- 14) \pm \sqrt {(- 14) ^ 2-4 (15) (3)}} {2 (15)}\\x = \frac {14 \pm \sqrt {196-180}} {30}\\x = \frac {14 \pm \sqrt {16}} {30}\\x = \frac {14 \pm4} {30}

So, we have:

x_ {1} = \frac{14 + 4} {30} = \frac {18} {30} = \frac {18} {30} = \frac {3} {5}\\x_ {2} = \frac {14-4} {30} = \frac {10} {30} = \frac {1} {3}\\

Answer:

x_ {1} = \frac {3} {5}\\x_ {2} = \frac {1} {3}

Option A

Option E

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Answer:

The maximum value of f is 363, which is reached in (11,11) and (-11,-11) and the minimum value of f is -33, which is reached in (√11,-√11) and (-√11,√11)

Step-by-step explanation:

f(x,y) = 3xy, lets find the gradient of f. First lets compute the derivate of f in terms of x, thinking of y like a constant.

f_x(x,y) = 3y

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f_y(x,y) = 3x

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\nabla{f} = (3y,3x)

The restriction is given by g(x,y) = 121, with g(x,y) = 3x²+3y²-5xy. The partial derivates of g are

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\nabla g(x,y) = (6x-5y,6y-5x)

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  • f_x(x_0,y_0) = \lambda \, g_x(x0,y0)
  • f_y(x_0,y_0) = \lambda \, g_y(x_0,y_0)
  • g(x_0,y_0) = 121

This can be translated into

  • 3y = \lambda (6x-5y)
  • 3x = \lambda (-5x+6y)
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Thus, x = -y or λ=3.

If x were -y, then we can replace x for -y in both equations

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Note that y cant take the value 0 because, since x = -y, we have that x = y = y, and g(x,y) = 0. Therefore, equation 3 wouldnt hold.

Now, lets suppose that λ=3, if that is the case, we can replace in the first 2 equations obtaining

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Lets evaluate g in (-y,y) and try to find y

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The candidates to extremes are, as a result (√11,-√11), (-√11, √11). In both cases, f(x,y) = 3 √11 (-√11) = -33

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In both cases f(11,11) = f(-11,-11) = 363.

We conclude that the maximum value of f is 363, which is reached in (11,11) and (-11,-11) and the minimum value of f is -33, which is reached in (√11,-√11) and (-√11,√11)

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