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.

Step-by-step 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.

`(2x^2)/(2)+(6x)/(2)+(2)/(2)=(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=(-b\pm√(b^2-4ac))/(2a)`

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

`x=(-3\pm√(3^2-4(1)(1)))/(2(1))`
`x=(-3\pm√(9-4))/(2)`
`x=(-3\pm√(5))/(2)`
`x=(-3\pm√(5))/(2)`