How do you find the roots, real and imaginary, of #y=4x^2 +x -3-(x-2)^2 # using the quadratic formula?

1 Answer
Jan 23, 2016

Answer:

#x=0.9067 and x=-2.5734#

Explanation:

first, expand the bracket

#(x-2)^2#

#(x-2)(x-2)#

#x^2-4x+4#

then, solve the equations

#y=4x^2+ x-3-(x^2-4x+4)#

#y=4x^2+ x-3-x^2+4x-4#

#y=3x^2+ 5x-7#

then, by using #b^2-4ac#

for the equation: #y=3x^2+ 5x-7#

where #a=3, b=5 and c=-7# into #b^2-4ac#

#5^2-4(3)(-7)#

#25--84#

#109#

so, compare to this

#b^2-4ac>0# : two real and different roots
#b^2-4ac=0# : two real root and equals
#b^2-4ac<0# : no real roots or (the roots are complexes)

so, #109>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= (-5+- sqrt(5^2-4(3)(-7) ))/ (2(3)#

#x= (-5+- sqrt(109))/ 6#

#x= (-5+sqrt(109))/ 6# and #x= (-5- sqrt(109))/ 6#

solve it and u will get the values of x which is

#x=0.9067 and x=-2.5734#