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loris [4]
2 years ago
14

Which of the following is the discriminant of the polynomial below x^2+4x+5

Mathematics
1 answer:
steposvetlana [31]2 years ago
4 0

Answer:

\huge\boxed{\sf Discriminant = -4}

Step-by-step explanation:

Comparing it with quadratic equation ax^2 + bx + c = 0, we get:

a = 1 . b = 4 and c = 5

So,

Discriminant = b^2-4ac

D = (4)²-4(1)(5)

D = 16 - 20

D = -4

Hence,

Discriminant = -4

\rule[225]{225}{2}

Hope this helped!

<h3 /><h3>~AH1807</h3>
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It means that the variables cancelled out and you might have made an error in your working, or it could also mean that that solution was not going to work out (hence leading 0=0). It all depends on the question. 

Hope I helped :) 
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What is the square root of -16?<br> 0-81<br> 041<br> OO
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4 i

Step-by-step explanation:

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consider the function and then use calculus to answer the questions that follow 1 1/x 5/x^2 1/x^3 (a) Find the interval(s) where
boyakko [2]

Answer:

a)X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

b)Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

Step-by-step explanation:

From the question we are told that

The Function

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

Generally the differentiation of function f(x) is mathematically solved as

f(x)=1+\frac{1}{x}  +\frac{5}{x^2} +\frac{1}{x^3}

f(x)=\frac{x^3+x^2+5x+1}{x^2}

Therefore

f'(x)=\frac{x^2+10x+3}{x^4}

Generally critical point is given as

f'(x)=0

\frac{x^2+10x+3}{x^4}=0

x=-5 \pm\sqrt{22}

Generally the maximum and minimum x value for critical point is mathematically solved as

f'(-5 \pm\sqrt{22})

Where

Maximum value of x

f'(-5 +\sqrt{22})

Minimum value of x

f'(-5 +\sqrt{22})

Therefore interval of increase is mathematically given by

f'(-5 -\sqrt{22}),f'(-5 +\sqrt{22})

f(x)

Therefore interval of decrease is mathematically given by

(-\infty,-5 -\sqrt{22}),f'(-5 +\sqrt{22},0),(0,\infty)

Generally the second differentiation of function f(x) is mathematically solved as

f''(x)=\frac{2(x^2+15x+6)}{x^5}

Generally the point of inflection is mathematically solved as

f''(x)=0

x^2+15x+6=0

Therefore inflection points is given as

x=\frac{1}{2} (-15 \pm \sqrt{201}

f''(x)>0,\frac{1}{2}(-15-\sqrt{201})

a)Generally the concave upward interval X is mathematically given as

X=((-15-\sqrt{201},(-15+\sqrt{201}),(0,\infty)

f''(x)

b)Generally the concave downward interval Y is mathematically given as

Y=(\infty,\frac{1}{2}(-15-\sqrt{201} ) ),(\frac{1}{2}()-15+\sqrt{201)},0  )

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Answer:

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Step-by-step explanation:

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