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3 years ago

Find the real solutions of the equation by graphing. x2 + 2x + 2 = 0

Mathematics

1 answer:

Natalija [7]3 years ago

8 0

Answer:

There is no real solution

Step-by-step explanation:

∵ x² + 2x + 2 = 0

From the graph there is no intersection between the parabola and

the x-axis

∴ There is no real solution

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How many solutions does this system have?

densk [106]

Answer:

One solution since x=0

6 0

3 years ago

Part one: How do you solve a system of equations approximately using graphs and tables?

tatiyna

This problem can be solve by graphing (technology), as suggested.
The answer is posted as an attached image. We see that after about 14.75 years, the invading species will surpass the indigenous population.

If it needs to be solved mathematically and accurately, the math is a little more advanced, using the bisection method, or Newton's method.
However, we can also do that by trial and error, starting from 14.75. It is easier than you might think.

Post if you would like to have more information on one or the other methods.

Note: the scale of y has been shrunk by 1000, so each unit on the y-axis represents 1000 frogs.

5 0

3 years ago

viva [34]

Answer:

(x - 1) (x + 2) (x + 4)

Step-by-step explanation:

Factor the following:

x^3 + 5 x^2 + 2 x - 8

The possible rational roots of x^3 + 5 x^2 + 2 x - 8 are x = ± 1, x = ± 2, x = ± 4, x = ± 8. Of these, x = 1, x = -2 and x = -4 are roots. This gives x - 1, x + 2 and x + 4 as all factors:

Answer: (x - 1) (x + 2) (x + 4)

6 0

4 years ago

Read 2 more answers

For integers a, b, and c, consider the linear Diophantine equation ax C by D c: Suppose integers x0 and y0 satisfy the equation;

Dmitrij [34]

Answer:

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

b. x = -8 and y = 4

Step-by-step explanation:

This question is incomplete. I will type the complete question below before giving my solution.

For integers a, b, c, consider the linear Diophantine equation

$ax+by=c$

Suppose integers x0 and yo satisfy the equation; that is,

$ax_0+by_0 = c$

what other values

$x = x_0+h$ and $y=y_0+k$

also satisfy ax + by = c? Formulate a conjecture that answers this question.

Devise some numerical examples to ground your exploration. For example, 6(-3) + 15*2 = 12.

Can you find other integers x and y such that 6x + 15y = 12?

How many other pairs of integers x and y can you find ?

Can you find infinitely many other solutions?

From the Extended Euclidean Algorithm, given any integers a and b, integers s and t can be found such that

$as+bt=gcd(a,b)$

the numbers s and t are not unique, but you only need one pair. Once s and t are found, since we are assuming that gcd(a,b) divides c, there exists an integer k such that gcd(a,b)k = c.

Multiplying as + bt = gcd(a,b) through by k you get

$a(sk) + b(tk) = gcd(a,b)k = c$

So this gives one solution, with x = sk and y = tk.

Now assuming that ax1 + by1 = c is a solution, and ax + by = c is some other solution. Taking the difference between the two, we get

$a(x_1-x) + b(y_1-y)=0$

Therefore,

$a(x_1-x) = b(y-y_1)$

This means that a divides b(y−y1), and therefore a/gcd(a,b) divides y−y1. Hence,

$y = y_1+r(\frac{a}{gcd(a, b)})$ for some integer r. Substituting into the equation

$a(x_1-x)=rb(\frac{a}{gcd(a, b)} )\\gcd(a, b)*a(x_1-x)=rba$

$x = x_1-r(\frac{b}{gcd(a, b)} )$

Thus if ax1 + by1 = c is any solution, then all solutions are of the form

$x = x_1+r(\frac{b}{gcd(a, b)} )\\y=y_1-r(\frac{a}{gcd(a, b)} )$

In order to find all integer solutions to 6x + 15y = 12

we first use the Euclidean algorithm to find gcd(15,6); the parenthetical equation is how we will use this equality after we complete the computation.

$15 = 6*2+3\\6=3*2+0$