How do you solve #y=-4x^2-x-3, y=x^2+2x-5#?
1 Answer
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]}