How do you solve y=-4x^2-x-3, y=x^2+2x-5y=−4x2−x−3,y=x2+2x−5?
1 Answer
Explanation:
color(red)(y)=-4x^2-x-3to(1)y=−4x2−x−3→(1)
color(red)(y)=x^2+2x-5to(2)y=x2+2x−5→(2) Since both equations are expressed in terms of y we can equate the right sides.
rArrx^2+2x-5=-4x^2-x-3⇒x2+2x−5=−4x2−x−3
"collect terms on left side and equate to zero"collect terms on left side and equate to zero
rArr5x^2+3x-2=0⇒5x2+3x−2=0
rArr(5x-2)(x+1)=0⇒(5x−2)(x+1)=0
rArrx=2/5" or " x=-1⇒x=25 or x=−1 Substitute these values into either ( 1 ) or ( 2 ) and evaluate for y
"evaluating in " (2)evaluating in (2)
x=2/5toy=(2/5)^2+2(2/5)-5=-101/25x=25→y=(25)2+2(25)−5=−10125
x=-1to(-1)^2+2(-1)-5=-6x=−1→(−1)2+2(−1)−5=−6
"points of intersection are " (2/5,-101/25),(-1,-6)points of intersection are (25,−10125),(−1,−6)
graph{(y+4x^2+x+3)(y-x^2-2x+5)=0 [-12.48, 12.49, -7.24, 6.24]}
