Question

Select the correct answer. The solution to this system of equations lies between the x-values -2 and -1.5. At which x-value are the two equations approximately equal? y=1/x+2 y=x^2+2

Answer 1

Answer:

The two equations will be equal approximately at x = -1.

Step-by-step explanation:

The given equations are $y = \frac{1}{x + 2} \\ y = x^{2} + 2$.

The equations will be approximately equal, when the equations intersect with each other. In order to get the intersecting points, we need to solve the equations.

Equating the two values of y, we get $\frac{1}{x + 2} = x^{2} + 2\\x^{3} + 2x^{2} + 2x + 2 = 1\\x^{3} + 2x^{2} + 2x + 1 = 0\\(x + 1)(x^{2} + x + 1) = 0.\\x = -1$.

There is no real value of x, so that $x^{2} + x + 1 = 0$.