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pantera1 [17]
3 years ago
6

Julie walked less than 2 miles? its 6 grade lesson8.3p

Mathematics
1 answer:
Novosadov [1.4K]3 years ago
6 0

Answer:

huh?, give the full answer please

Step-by-step explanation:

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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
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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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Step-by-step explanation:

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The product of 7 and the sum of a number and -6
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