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AysviL [449]
3 years ago
6

If c and k are the roots of 6-x-x^2=0 find c + k. pls solve like a quadratic equation​

Mathematics
2 answers:
beks73 [17]3 years ago
4 0

Answer:

c + k = - 1

Step-by-step explanation:

given a quadratic equation in standard form : ax² + bx + c = 0 : a ≠ 0

Then the sum of the roots = - \frac{b}{a} and

The product of the roots = \frac{c}{a}

- x² - x + 6 = 0 is in standard form

with a = - 1, b = - 1 and c = 6

The roots are c and k, hence

c + k = - \frac{-1}{-1} = - 1

Ratling [72]3 years ago
3 0

by Viete's formula

c + k = - koef x/koef x²

= - (-1)/(-1) = -1

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Step-by-step explanation:

1) √4 . √-3 . √-3

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2) √-4 . √-3 . √-3

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3) √4 . √3 . √-3

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$ \sqrt{3} . \sqrt{-3} = (\sqrt{3})^2 . \sqrt{-1} $

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$ \sqrt{3} . \sqrt{3} = (\sqrt{3})^2  = 3 $

Therefore, √4 . √3 . √3 = 2 . 3 = 6

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To solve 10x-35+3ax=5ax-7a, we'll need to assume that a is a constant and that we are to solve for the variable x.  To accomplish this, group all the x terms together on one side and all of the no-x terms on the other:

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