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DanielleElmas [232]
2 years ago
14

The zeroes of the polynomial f(x) = x^2+x+3/4 are​

Mathematics
2 answers:
matrenka [14]2 years ago
6 0

Step-by-step explanation:

f(x) = x² + x + 3/4

in general, such a quadratic function is defined as

f(x) = a×x² + b×x + c

the solution for finding the values of x where a quadratic function value is 0 (there are as many solutions as the highest exponent of x, so 2 here in our case)

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = 1

c = 3/4

x = (-1 ± sqrt(1² - 4×1×3/4))/(2×1) =

= (-1 ± sqrt(1 - 3))/2 = (-1 ± sqrt(-2))/2 =

= (-1 ± sqrt(2)i)/2

x1 = (-1 + sqrt(2)i) / 2

x2 = (-1 - sqrt(2)i) / 2

remember, i = sqrt(-1)

f(x) has no 0 results for x = real numbers.

for the solution we need to use imaginary numbers.

jeka942 years ago
4 0

Answer:

  • No real zeros

Step-by-step explanation:

<u>I'll solve it by completing the square:</u>

f(x) = x² + x + 3/4 =

        x² + 2*(1/2)x + (1/2)² - (1/2)² + 3/4 =

        (x + 1/2)² - 1/4 + 3/4 =

        (x + 1/2)² + 1/2

We got the expression, which is greater than zero for any value of x. It means the graph has no intersection with the x-axis, and there are no real zero's.

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aliya0001 [1]

Answer:

The closest points on the cone are;

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Step-by-step explanation:

Let B(x, y, z) denote a point on the cone.

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d = √[(x - 6)² + (y - 2)² + (z - 0)²]

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Thus, the closest points on the cone are;

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