first, expand the bracket
(x-2)^2(x−2)2
(x-2)(x-2)(x−2)(x−2)
x^2-4x+4x2−4x+4
then, solve the equations
y=4x^2+ x-3-(x^2-4x+4)y=4x2+x−3−(x2−4x+4)
y=4x^2+ x-3-x^2+4x-4y=4x2+x−3−x2+4x−4
y=3x^2+ 5x-7y=3x2+5x−7
then, by using b^2-4acb2−4ac
for the equation: y=3x^2+ 5x-7y=3x2+5x−7
where a=3, b=5 and c=-7a=3,b=5andc=−7 into b^2-4acb2−4ac
5^2-4(3)(-7)52−4(3)(−7)
25--8425−−84
109109
so, compare to this
b^2-4ac>0b2−4ac>0 : two real and different roots
b^2-4ac=0b2−4ac=0 : two real root and equals
b^2-4ac<0b2−4ac<0 : no real roots or (the roots are complexes)
so, 109>0109>0 means two real and different roots
thus, you must use this formula to find the imaginary roots
x= (-b+- sqrt(b^2-4ac ))/ (2a)x=−b±√b2−4ac2a
x= (-5+- sqrt(5^2-4(3)(-7) ))/ (2(3)x=−5±√52−4(3)(−7)2(3)
x= (-5+- sqrt(109))/ 6x=−5±√1096
x= (-5+sqrt(109))/ 6x=−5+√1096 and x= (-5- sqrt(109))/ 6x=−5−√1096
solve it and u will get the values of x which is
x=0.9067 and x=-2.5734x=0.9067andx=−2.5734
