How do you solve #-16x ^ { 2} = 12x ^ { 2} + 24x + 5#?

2 Answers
Jan 15, 2018

Answer:

#x = -5/14, -1/2#

Explanation:

In order to solve this equation, first we will have to convert it into the general form, #ax^2 + bx + c = 0#, where #a != 0#.

So, #-16x^2 = 12x^2 + 24x + 5#

#rArr 28x^2 + 24x + 5 = 0# [Transposing #-16x^2# to the R.H.S]

Now, we will use the Quadratic Formula or Sridhar Acharya's Formula, to solve for the two roots.

Here, Discriminant = #D# = #b^2 - 4ac = (24)^2 - 4 * 28 * 5#

#= 576 - 560 = 16 gt 0#

So, the equation will have two real and distinct roots.

Now, Applying the Formula,

#alpha# = #(- b + sqrt(D))/(2a) = (- 24 + sqrt(16))/(2 * 28) = -20/56 = -5/14#

and, #beta# = #(-b - sqrt(D))/(2a) = (- 24 - 4)/(2 * 28) = -28/56 = -1/2#

So, the two roots of the equation are #-5/14# and #-1/2#.

Answer:

See a solution process below:

Explanation:

First, add #color(red)(16x^2)# to each side of the equation to put the equation in standard form:

#-16x^2 + color(red)(16x^2) = 12x^2 + color(red)(16x^2) + 24x + 5#

#0 = (12 + color(red)(16))x^2 + 24x + 5#

# 0 = 28x^2 + 24x + 5#

#28x^2 + 24x + 5 = 0#

Next. we can factor the left side of the equation as:

#(14x + 5)(2x + 1) = 0#

Now, we can solve each term on the left for #0#:

Solution 1:

#14x + 5 = 0#

#14x + 5 - color(red)(5) = 0 - color(red)(5)#

#14x + 0 = -5#

#14x = -5#

#(14x)/color(red)(14) = -5/color(red)(14)#

#(color(red)(cancel(color(black)(14)))x)/cancel(color(red)(14)) = -5/14#

#x = -5/14#

Solution 2:

#2x + 1 = 0#

#2x + 1 - color(red)(1) = 0 - color(red)(1)#

#2x + 0 = -1#

#2x = -1#

#(2x)/color(red)(2) = -1/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -1/2#

#x = -1/2#

The Solution Is:

#x = {-1/2, -5/14}#