How many solutions does this system of linear equations have? Explain. 1= 2 =1+3 2

Mathematics

1 answer:

Neporo4naja [7]3 years ago

8 0

Answer: There are no real solutions.

Step-by-step explanation:

We have the equation:

x = y^2

y = x + 3/2

To solve this, we can start by noticing that x is already isolated in the first equation, then we can replace it in the second equation to get:

y = y^2 + 3/2

And now we can rewrite this as:

y^2 - y + 3/2 = 0

This is a quadratic equation, and the solutions can be calculated with Bhaskara's equation, the solutions are:

$y = \frac{-(-1) +- \sqrt{(-1)^2 - 4*3/2*1} }{2*1} = \frac{1 + \sqrt{-5} }{2}$

We have a negative number inside the square root, this means that the solutions are complex numbers.

Then we can conclude that the system of equations has no real solutions.

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Which of the following definite integrals could be used to calculate the total area bounded by the graph of y = x^3 – 3x^2 + 2x

castortr0y [4]

Answer: The correct option is second, i.e. ,"2 times the integral from 0 to 1 of the quantity x cubed minus 3 times x squared plus 2 times x, dx".

Explanation:

The given equation is,

$y=x^3-3x^2+2x$

It can be written as,

$f(x)=x^3-3x^2+2x$

Find the zeros of the equation. Equation the function equal to 0.

$0=x^3-3x^2+2x$

$x(x^2-3x+2)=0$

$x(x^2-2x-x+2)=0$

$x(x-2)(x-1)=0$

So, the three zeros are 0, 1 and 2.

The graph of the equation is shown below.

From the given graph it is noticed that the enclosed by the curve and x- axis is lies between 0 to 2, but the area from 0 to 1 lies above the x-axis and area from 1 to 2 lies below the x-axis. So the function will be negative from 1 to 2.

The area enclosed by curve and x-axis is,

$A=\int_{0}^{1}f(x)dx+\int_{1}^{2}[-f(x)]dx$