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Vika [28.1K]
3 years ago
15

What does k^2 -8k-20=0

Mathematics
2 answers:
Karolina [17]3 years ago
7 0
k^{2}-8k-20=0 \\ (k-10)(k+2)=0 \\  \\ k-10=0 \\ k=10 \\  \\ k+2=0 \\ k=-2 \\  \\ \boxed{k=-2\ and\ 10}
victus00 [196]3 years ago
3 0
K² - 8k - 20 = 0
k = <u>-(-8) +/- √((-8)² - 4(1)(-20))</u>
                       2(1)
k = <u>8 +/- √(64 + 80)</u>
                 2
k = <u>8 +/- √(144)
</u>              2<u>
</u>k = <u>8 +/- 12</u><u>
</u>            2
k = <u>8 + 12</u>          k = <u>8 - 12</u>
          2                        2
h = <u>20</u>                k = <u>-4</u>
       2                        2
x = 10                k = -2
<u />
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Answer:

\large\boxed{(x-2)^2+(y-1)^2=34}

Step-by-step explanation:

The equation of a circle in standard form:

(x-h)^2+(y-k)^2=r^2

(h, k) - center

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Midpoint of diameter is a center of a circle.

The formula of a midpoint:

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Substitute:

h=\dfrac{-1+5}{2}=\dfrac{4}{2}=2\\\\k=\dfrac{6+(-4)}{2}=\dfrac{2}{2}=1

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The radius length is equal to the distance between the center of the circle and the endpoint of the diameter.

The formula of a distance between two points:

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Substitute the coordinates of the points (2, 1) and (5, -4):

r=\sqrt{(5-2)^2+(-4-1)^2}=\sqrt{3^2+(-5)^2}=\sqrt{9+25}=\sqrt{34}

Finally we have:

(x-2)^2+(y-1)^2=(\sqrt{34})^2

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