How do you find the roots, real and imaginary, of #y= (2x-1)^2+2 (x + 1) (x - 4) # using the quadratic formula?

2 Answers
Feb 10, 2018

Answer:

#y = (5+sqrt(79))/6# and #(5-sqrt(79))/6#

Explanation:

First things first, we need to get this into standard form, #ax^2+bx+c#

expand

#(2x-1)(2x-1)+2(x+1)(x-4)#

distribute

#4x^2-4x-1 + (2x+2)(x-4)#

#4x^2-4x-1 + 2x^2-6x-8#

combine like-terms

#color(orange)(6)x^2 + color(blue)(-10)x + color(pink)(-9)#

The quadratic formula is #(-b)/(2a) +- ( sqrt( b^2 - 4 xx a xx c ) )/(2a)#

with #a = color(orange)(6)#, #b = color(blue)(-10)#, and #c = color(pink)(-9)#

#(-color(blue)(-10))/(2 xx color(orange)(6)) +- ( sqrt( (color(blue)(-10))^2 - 4 xx color(orange)(6) xx color(pink)(-9) ) )/(2 xx color(orange)(6))#

simplify

#10/12 +- ( sqrt( 100 + 216 ) )/(12)#

#5/6 +- (2sqrt79)/12#

#5/6 +- sqrt(79)/6#

#y = (5+sqrt(79))/6# and #(5-sqrt(79))/6#

Feb 10, 2018

Answer:

#y= f(x) = (2x-1)^2+2 (x + 1) (x - 4) " " # (or)

#color(blue)(y=6x^2-10x-7#

Roots : #color(blue)(x_1 = [5-sqrt(67)]/6, " " x_2 = [5+sqrt(67)]/6 " "# (or)

#x_1 = -0.53089, or x_2 = 2.19756#

Explanation:

Given:

#y= f(x) = (2x-1)^2+2 (x + 1) (x - 4) " " # Polynomial

#color(green)(Step.1#

We will expand and simplify

#y= f(x) = (2x-1)^2+2 (x + 1) (x - 4) #

as shown below:

First, consider #color(blue)((2x-1)^2#

Using the algebraic identify,

#color(brown)((a-b)^2 -= a^2 - 2ab - b^2#,

we can expand as shown:

#rArr (2x)^2 - 2(2x)*(1) + (1)^2#

#rArr 4x^2-4x+1 " "# Expression.1

#color(green)(Step.2#

Next, consider

#color(blue)(2 (x + 1) (x - 4) #

We can factor them out to obtain

#rArr 2*(x^2 - 4x+x-4)#

#rArr 2*(x^2 - 3x-4)#

#rArr 2x^2 - 6x - 8 " "# Expression.2

#color(green)(Step.3#

We will add our Expression.1 and Expression.2 to get #y#

# (4x^2-4x+1)+(2x^2 - 6x - 8)#

#rArr 4x^2-4x+1+2x^2-6x-8#

#rArr 6x^2-10x-7#

Now, we are in a position to re-write our Polynomial as

#color(blue)(y=f(x)=6x^2-10x-7#

#color(blue)(6x^2-10x-7=0# is our Quadratic Equation,

and we proceed to find the roots.

#color(green)(Step.4#

Now that we have a Quadratic Equation in the Standard Form

#ax^2 + bx + c = 0#, where

#a = 6; b = (-10) and c = (-7)#

We now use the Quadratic Formula to find the roots

#color(red)(Root_1,_2 = -b+- sqrt(b^2-4*a*c)/(2.a)#

#rArr [-(-10)+-sqrt((-10)^2-4(6)(-7))]/(2(6)#

#rArr [10+-sqrt(100-4(-42)]]/12#

#rArr [10+-sqrt(100+168]]/12#

#rArr [10+-sqrt(268]]/12#

#rArr 10/12+-sqrt(268)/12#

#rArr 10/12+-sqrt(4*67)/12#

#rArr 10/12+-sqrt(4)/12*sqrt(67)/12#

#rArr 5/6+-2/12*sqrt(67)/12#

#rArr 5/6+-(2sqrt(67))/12#

#rArr 5/6+-(cancel 2sqrt(67))/(cancel 12( color(red)(6))#

#rArr 5/6 +-sqrt(67)/6#

#rArr (5+-sqrt(67))/6#

Hence, we can write our roots as

#color(blue)(Root_1 = [5-sqrt(67)]/6, " " Root_2 = [5+sqrt(67)]/6 " "# (or)

#Root_1 = -0.53089, or Root_2 = 2.19756#

Please observe that these are our required real roots, as the discriminant #color(blue)((b^2-4*a*c)>0#

#color(green)(Step.5#

We will verify our solutions using GeoGebra graphing software

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