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3 years ago

C²+5c-24=0 cant figure out how to solve this

Mathematics

2 answers:

Bumek [7]3 years ago

5 0

Move the 24 to the other side and that's the farthest it simplifies

frosja888 [35]3 years ago

4 0

Just solve the quadratic, you need to figure out what value you can use to FOIL and get c^2+5c-24=0. so you need to figure out what values you can add together to get 5 and multiply together to get -24 so try (c+8)(c-3) here 8 and 3 make sense since 8-3 = 5 and 8*-3=-24 so the answer is (c+8)(c-3).

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Step-by-step explanation:

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zaharov [31]

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(y-p)/2

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The roots of the equation are 3 + 8i and 3 – 8i.

g100num [7]

Given that the roots of the equation x^2-6x+c=0 are 3+8i and 3-8i, the value of c can be obtained as follows;
taking x=3+8i and substituting it in our equation we get:
(3+8i)^2-6(3+8i)+c=0
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3 0

3 years ago

6. Line a: y - 3x = 4<br> Line b: y = 3x + 5<br> Line c: y - 4x = 3<br><br> Which lines are parallel

kramer

Answer:

Line a and b

Step-by-step explanation:

6 0

1 year ago

The circle given by : x^2 + y^2 - 6y - 12 = 0 can be written as:

deff fn [24]

Step-by-step explanation:

Given equation of circle is,

${x}^{2} + {y}^{2} - 6y - 12 = 0 \\ by \: compairing \: it \: with \: \\ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: we \: get \\2 g = 0 \: or \: g = 0 \\ 2f= - 6 \\ or \: f= - 3 \\ c = - 12$

again the another form of circle is,

${x}^{2} + {(y - k)}^{2} = 21 \\ or \: {x}^{2} + {y}^{2} - 2ky + {k}^{2} - 21 = 0 \\ by \: compairing \:it \: with \: \\ {x}^{2} + {y}^{2} + 2gx + 2fy = 0 \: we \: get \\ g = 0 \\f = - k \\ c = {k}^{2} - 21$

now equating the values of f in both equations,

-k=-3

i.e. k=3

therefore k=3

4 0

2 years ago