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Kisachek [45]
3 years ago
7

0,4 Find the roots of the following quadratic equations

Mathematics
1 answer:
Crazy boy [7]3 years ago
7 0

Answer:

b = b2-4ac

= (-7ab)^2 -4 (6a^2)(-3b^2)

= 49a^2b^2 + 72a^2b^2

= 121a^2b^2

Now, x = (7ab +- root 121a^2b^2)÷2(6a²)

= (7ab +- 11ab ) ÷ 12a²

= ( 20ab)/12a² or 4ab/12a²

= 5/3ab or - 1/3ab .

So, these are its roots .

Step-by-step explanation:

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Step-by-step explanation:

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8 0
3 years ago
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Step-by-step explanation:

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((15t-8)(4-t))/5 < -7

(15t^2)-28t-32< -7

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Using the quadratic/almighty formula

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Since

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8 0
3 years ago
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forsale [732]

Answer:

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So the point where it scoops up is (1, -18)

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The places where it touches the x-axis is -1 and 3

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