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saveliy_v [14]
3 years ago
13

For what real value of $v$ is $\frac{-21-\sqrt{301}}{10}$ a root of $5x^2+21x+v$?

Mathematics
1 answer:
Degger [83]3 years ago
6 0

Answer:

v = 7

is the value for which

x = (-21 - √301)/10

is a solution to the quadratic equation

5x² + 21x + v = 0

Step-by-step explanation:

Given that

x = (-21 - √301)/10 .....................(1)

is a root of the quadratic equation

5x² + 21x + v = 0 ........................(2)

We want to find the value of v foe which the equation is true.

Consider the quadratic formula

x = [-b ± √(b² - 4av)]/2a ..................(3)

Comparing (3) with (2), notice that

b = 21

2a = 10

=> a = 10/2 = 5

and

b² - 4av = 301

=> 21² - 4(5)v = 301

-20v = 301 - 441

-20v = -140

v = -140/(-20)

v = 7

That is a = 5, b = 21, and v = 7

The equation is then

5x² + 21x + 7 = 0

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Answer:
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Complete the statements about the key features of the graph of f(x) = x5 – 9x3. As x goes to negative infinity, f(x) goes to neg
Mars2501 [29]
The function is f(x)= x^{5} -9x ^{3}

1. let's factorize the expression x^{5} -9x ^{3}:

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the zeros of f(x) are the values of x which make f(x) = 0.

from the factorized form of the function, we see that the roots are:

-3, multiplicity 1
3, multiplicity 1
0, multiplicity 3

(the multiplicity of the roots is the power of each factor of f(x) )


2.  
The end behavior of f(x), whose term of largest degree is x^{5}, is the same as the end behavior of  x^{3}, which has a well known graph. Check the picture attached. 

(similarly the end behavior of an even degree polynomial, could be compared to the end behavior of x^{2})

so, like the graph of x^{3}, the graph of f(x)= x^{5} -9x ^{3} :

"As x goes to negative infinity, f(x) goes to negative infinity, and as x goes to positive infinity, f(x) goes to positive infinity. "

7 0
3 years ago
Read 2 more answers
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