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Artyom0805 [142]
3 years ago
7

Solve 9x^2 + 42x + 49 = 0

Mathematics
1 answer:
svet-max [94.6K]3 years ago
3 0

Because of the relatively large coefficients {9, 42, 49}, applying the quadratic formula would be a bit messy. Instead, I've chosen to "complete the square:"


9x^2 + 42x + 49 = 0 can be re-written as 9 [ x^2 + (42/9)x ] = -49


Dividing both sides by 9, we get [ x^2 + (42/9)x ] = - 49/9


Completing the square: [ x^2 + (42/9)x + (21/9)^2 - (21/9)^2 ] = -49/9

[ x + 21/9 ]^2 = 441/81 - 441/81 = 0


Then [ x + 21/9 ] = 0, and x = -21/9 (this is a double root).

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So the number of pencils he brought is 13.

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Answer: Vertex = (2, -15)  2nd point = (0, -3)

<u>Step-by-step explanation:</u>

g(x) = 3x² - 12x - 3

      = 3(x² - 4x - 1)

          a=1   b=-4  c=-1

Find the x-value of the vertex by using the formula for the axis of symmetry: x = \dfrac{-b}{2a}

x = \dfrac{-(-4)}{2(1)}

      = \dfrac{4}{2}

         = 2

Find the y-value of the vertex by plugging the x-value (above) into the given equation: g(x) = 3x² - 12x - 3

g(2) = 3(2)² - 12(2) - 3

       = 12  - 24 - 3

       = -15

So, the vertex is (2, -15)  ←  PLOT THIS COORDINATE

Now, choose a different x-value.  Plug it into the equation and solve for y. <em>I chose x = 0</em>

g(0) = 3(0)² - 12(0) - 3

       = 0  - 0 - 3

       = -3

So, an additional point is (0, -3)  ←  PLOT THIS COORDINATE


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3 years ago
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