Question

(x^2 -x)^2 +5(x^2-x) +4 =0
solve this quadratic equation ​

Answer 1

Step-by-step explanation:

let the x^2-x=y

then,

y^2+5y+4=0

y^2+4y+y+4=0

y(y+4)+1(y+4)=0

(y+4)(y+1)=0

y=-4

y=-1

case :y=-4)

x^2-x=-4

x^2-x+4=0

x=1+√(1-16)/2. or x=1-√(1-16)/2

x=1+√(-15)/2. or x=1-/√(-15)/2

case: y=-1)

x^2-x=-1

x^2-x+1=0

x=1+√(1-4)/2. or x=1-√(1-4)/2

x=1+√(-3)/2. or x=1-√(-3)/2