How do you simplify #(9x^2-16)/ (6x^2-11x+4) div ( 6x^2+11x+4)/(8x^2+10x+3)#?

1 Answer
Jul 30, 2015

Answer:

#E = [9 * (x + 3/4)]/((x-1/2))#

Explanation:

So, your starting expression looks like this

#E = (9x^2 - 16)/(6x^2 - 11x + 4) * (8x^2 + 10x + 3)/(6x^2 + 11x + 4)#

You can use the quadratic formula to factor the denominators of the two fractions, plus the numerator of the second one.

For the general form quadratic #ax^2 + bx + c = 0#, you have

#color(blue)(x_(1,2) = (-b +-sqrt(b^2 - 4ac))/(2a))#

You will get

#6x^2 - 11x + 4 = 0#

#x_(1,2) = (11 +- sqrt((-11)^2 - 4 * 6 * 4))/(2 * 6)#

#x_(1,2) = (11 +-5)/12 = {(x_1 = (11 + 5)/12 = 4/3), (x_2 = (11-5)/12 = 1/2) :}#

Notice that #6x^2 + 11 + 4 = 0# only differs from the previous quadratic by the sign of the #x#-term, which means that you can write

#x_(1,2) = (-11 +- 5)/12 = {(x_1 = (-11-5)/12 = -4/3), (x_2 = (-11 + 5)/12 = -1/2) :}#

FInally,

#8x^2 + 10x + 3 = 0#

#x_(1,2) = (-10 +- sqrt(10^2 - 4 * 8 * 3))/(2 * 8)#

#x_(1,2) = (-10 +- 2)/16 = { (x_1 = (-10 -2)/16 = -3/4), (x_2 = (-10 + 2)/16 = -1/2) :}#

Your original expression will now become

#E = (9x^2 - 16)/((x-1/2) * (x-4/3)) * (color(red)(cancel(color(black)((x + 1/2)))) * (x + 3/4))/(color(red)(cancel(color(black)((x + 1/2)))) * (x + 4/3))#

#E = (9x^2 - 16)/((x-1/2) * (x-4/3)) * (x+3/4)/(x + 4/3)#

Now focus on the numerator of the first fraction. Notice that if you use #9# as a common factor, you can write

#9 * (x^2 - 16/9)#

which is equivalent to

#9 * (x^2 - 16/9) = 9 * [x^2 - (4/3)^2] = 9 * (x-4/3) * (x + 4/3)#

Plug this into expression #E# to get

#E = (9 * color(red)(cancel(color(black)((x-4/3))) * color(red)(cancel(color(black)((x + 4/3))))))/( (x-1/2) * color(red)(cancel(color(black)((x -4/3))))) * (x + 3/4)/color(red)(cancel(color(black)(x + 4/3)))#

#E = color(green)((9 * (x + 3/4))/((x - 1/2)))#