Question

How do you solve `y=-4x^2-x-3, y=x^2+2x-5`?

Answer 1

`(2/5,-101/25),(-1,-6)`

Explanation:

`color(red)(y)=-4x^2-x-3to(1)`

`color(red)(y)=x^2+2x-5to(2)`

Since both equations are expressed in terms of y we can equate the right sides.

`rArrx^2+2x-5=-4x^2-x-3`

`"collect terms on left side and equate to zero"`

`rArr5x^2+3x-2=0`

`rArr(5x-2)(x+1)=0`

`rArrx=2/5" or " x=-1`

Substitute these values into either ( 1 ) or ( 2 ) and evaluate for y

`"evaluating in " (2)`

`x=2/5toy=(2/5)^2+2(2/5)-5=-101/25`

`x=-1to(-1)^2+2(-1)-5=-6`

`"points of intersection are " (2/5,-101/25),(-1,-6)`
graph{(y+4x^2+x+3)(y-x^2-2x+5)=0 [-12.48, 12.49, -7.24, 6.24]}