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BARSIC [14]
2 years ago
10

= 2. Given the quadratic equation 3x2 – 8x + k = 0 has no real roots. Find the range of values of k.

Mathematics
1 answer:
VashaNatasha [74]2 years ago
6 0

\text{If there  are no real roots, then}\\\\~~~~~\text{Discriminant} < 0\\\\\implies b^2 -4ac < 0\\\\\implies (-8)^2 -4 \cdot 3 k < 0\\\\\implies 64-12k < 0\\\\\ \implies 12k > 64\\\\\implies k > \dfrac{64}{12}\\\\\implies k > \dfrac{16}3\\\\\text{Interval,}~ \left( \dfrac{16}3, ~\infty\right)

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Answer:   (-2, 5) and (2, -3)

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

Graph the line y = -2x + 1 (which is in y = mx + b format) by plotting the y-intercept (b = 1) on the y-axis and then using the slope (m = -2) to plot the second point by going down 2 and right 1 unit from the first point:

        y - intercept = (0, 1)            2nd point = ( -1, 1).

Graph the parabola y = x² - 2x - 3 by first plotting the vertex and then plotting the y-intercept (or some other point):

y = x^2-2x-3\quad \rightarrow \quad a=1,\ b=-2,\ c=-3\\\\\text{axis of symmetry:}\ x = \dfrac{-b}{2a}\ \longrightarrow \ x=\dfrac{-(-2)}{2(1)}=\dfrac{2}{2}=1\\\\\text{y-value of vertex:}\ f(1) = (1)^2-2(1)-3\quad \longrightarrow \quad y = 1 - 2 - 3=-4\\\\\text{y-intercept:}\ f(0)= (0)^2-2(0)-3\ \longrightarrow \ y=0 - 0 - 3 = -3 \\

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<em>see attached</em> - the graphs intersect at two points:  (-2, 5) and (2, -3)


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3 years ago
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Answer:

A y =

1

2

x − 6

Step-by-step explanation:

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Read 2 more answers
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