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Dima020 [189]
4 years ago
9

What is the solution set of x2 + 5x + 1 = 0?

Mathematics
1 answer:
DerKrebs [107]4 years ago
3 0

Answer:

\huge\boxed{\left\{-\dfrac{5+\sqrt{21}}{2};\ \dfrac{-5+\sqrt{21}}{2}\right\}}

Step-by-step explanation:

x^2+5x+1=0\\\\\text{Use the quadratic formula:}\\\\\text{For}\ ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\\text{if}\ \Delta < 0,\ \text{then the equation has no real solution}\\\text{if}\ \Delta=0,\ \text{then the equation has one real solution}\ x=\dfrac{-b}{2a}\\\text{if}\ \Delta>0,\ \text{then the equation has two real solution}\ x=\dfrac{-b\pm\sqrt{\Delta}}{2a}

\text{We have}\\\\a=1,\ b=5,\ c=1\\\\\Delta=5^2-4(1)(1)=25-4=21>0\\\\x=\dfrac{-5\pm\sqrt{21}}{2(1)}=\dfrac{-5\pm\sqrt{21}}{2}

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Answer:

The minimum value of the given function is f(0) = 0

Step-by-step explanation:

Explanation:-

Extreme value :-  f(a, b) is said to be an extreme value of given function 'f' , if it is a maximum or minimum value.

i) the necessary and sufficient condition for f(x)  to have a maximum or minimum at given point.

ii)  find first derivative f^{l} (x) and equating zero

iii) solve and find 'x' values

iv) Find second derivative f^{ll}(x) >0 then find the minimum value at x=a

v) Find second derivative f^{ll}(x) then find the maximum value at x=a

Problem:-

Given function is f(x) = log ( x^2 +1)

<u>step1:</u>- find first derivative f^{l} (x) and equating zero

  f^{l}(x) = \frac{1}{x^2+1} \frac{d}{dx}(x^2+1)

f^{l}(x) = \frac{1}{x^2+1} (2x)  ……………(1)

f^{l}(x) = \frac{1}{x^2+1} (2x)=0

the point is x=0

<u>step2:-</u>

Again differentiating with respective to 'x', we get

f^{ll}(x)=\frac{x^2+1(2)-2x(2x)}{(x^2+1)^2}

on simplification , we get

f^{ll}(x) = \frac{-2x^2+2}{(x^2+1)^2}

put x= 0 we get f^{ll}(0) = \frac{2}{(1)^2}   > 0

f^{ll}(x) >0 then find the minimum value at x=0

<u>Final answer</u>:-

The minimum value of the given function is f(0) = 0

5 0
4 years ago
How would I answer this?
miss Akunina [59]

with the help of your calculator

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Answer:

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