Question

Solve the system of equationsy=-2x-3 y=2x^2+4x-1Round to the nearest hundredth

Answer 1

Given:

$y=-2x-3$$y=2x^2+4x-1$

Requires:

We need to solve the given system of equations.

Explanation:

Consider equation.

$y=2x^2+4x-1$$Substitute\text{ }y=-2x-3\text{ in the equation to find the value of x.}$$-2x-3=2x^2+4x-1$

Add 2x+3 to both sides of the equation.

$-2x-3+2x+3=2x^2+4x-1+2x+3$$0=2x^2+6x+2$$2x^2+6x+2=0$

Divide both sides of the equation by 2.

$\frac{2x^2}{2}+\frac{6x}{2}+\frac{2}{2}=\frac{0}{2}$$x^2+3x+1=0$

Which is of the form

$ax^2+bx+c=0$

where a=1, b=3, and c =1.

Consider the quadratic formula.

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Substitute a=1, b=3, and c =1 in the equaiton.

$x=\frac{-3\pm\sqrt{3^2-4(1)(1)}}{2(1)}$$x=\frac{-3\pm\sqrt{9-4}}{2}$$x=\frac{-3\pm\sqrt{5}}{2}$$x=\frac{-3\pm\sqrt{5}}{2}$