How do you find the roots, real and imaginary, of y=(2x-8)(x-2)-x^2-x y=(2x−8)(x−2)−x2−x using the quadratic formula?

1 Answer
Vikki
Dec 6, 2015

x~= 13.844 or x~= 1.156x≅13.844orx≅1.156

Explanation:

Step 1: Multiply/FOIL the expression

y= (2x-8)(x-2)-x^2 -xy=(2x−8)(x−2)−x2−x
y = 2x^2 -4x -8x + 16-x^2 -xy=2x2−4x−8x+16−x2−x
y = x^2 -15x + 16y=x2−15x+16

Step 2: Set the equation equal to zero
0= x^2 -15x + 160=x2−15x+16

Quadratic formula
x= (-b+-sqrt(b^2-4ac))/2ax=−b±√b2−4ac2a

a= 1 , b = -15, c= 16a=1,b=−15,c=16

Substitute into the formula

x = (-(-15) +- sqrt((-15)^2 -4(1)(16)))/2x=−(−15)±√(−15)2−4(1)(16)2

x = (15 +- sqrt(225 -64))/2x=15±√225−642

x= (15 +sqrt(161))/2x=15+√1612

x= (15+-(12.688577))/2x=15±(12.688577)2

x = (15+12.688577)/2 = 13.844288x=15+12.6885772=13.844288 or

x= (15- 12.688577)/2 = 1.1557115x=15−12.6885772=1.1557115

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