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stellarik [79]
3 years ago
13

show that thw roots of the equation (x-a)(x_b)=k^2 are always real if a,b and k are real. Please I really need help with this

Mathematics
1 answer:
VLD [36.1K]3 years ago
8 0

Answer:

see explanation

Step-by-step explanation:

Check the value of the discriminant

Δ = b² - 4ac

• If b² - 4ac > 0 then roots are real

• If b² - 4ac = 0 roots are real and equal

• If b² - 4ac < 0 then roots are not real

given (x - a)(x - b) = k² ( expand factors )

x² - bx - ax - k² = 0 ( in standard form )

x² + x(- a - b) - k² = 0

with a = 1, b = (- a - b), c = -k²

b² - 4ac = (- a - b)² + 4k²

For a, b, k ∈ R then (- a - b)² ≥ 0 and 4k² ≥ 0

Hence roots of the equation are always real for a, b, k ∈ R


           

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There are 256 squares, and you can count 16 on each edge. this shows 16 times 16, or 16 squared, which is 256.

3 0
3 years ago
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marta [7]
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3 years ago
In this polygon, all angles are right angles.
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The area of the given polygon is  1830 cm².

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5 0
1 year ago
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ANTONII [103]
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5 0
3 years ago
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kobusy [5.1K]

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Formula:

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r = I/(Pt)

r = 1350/(5000 x 3)

r = 1350/15000

r = 0.09

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5 0
2 years ago
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