How do you solve #(2x + 1) ( 4x ^ { 2} + 35x - 9) = 0#?

2 Answers
smendyka
Mar 16, 2017

See the entire solution process below:

Explanation:

First, factor the binomial to:

#(2x + 1)(4x - 1)(x + 9) = 0#

Now, solve each term for #0#:

Solution 1)

#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#

Solution 2)

#4x - 1 = 0#

#4x - 1 + color(red)(1) = 0 + color(red)(1)#

#4x + 0 = 1#

#4x = 1#

#(4x)/color(red)(4) = 1/color(red)(4)#

#(color(red)(cancel(color(black)(4)))x)/cancel(color(red)(4)) = 1/4#

#x = 1/4#

Solution 3)

#x + 9 = 0#

#x + 9 - color(red)(9) = 0 - color(red)(9)#

#x + 0 = -9#

#x = -9#

The solution is: #x = -1/2# and #x = 1/4# and #x = -9#

kamil9234
Mar 16, 2017

#x=-1/2#
#x=1/4#
#x=-9#

Explanation:

If any of paranthesises ( ) will be equal to 0, result will be 0 too.

#2x+1=0#
#2x=-1#
#x=-1/2#

OR

#4x^2+35x-9#
#(4x-1)(x+9)=0#

SO, from THIS we have:

#4x-1=0#
#4x=1#
#x=1/4#

AND

#x+9=0#
#x=-9#

THE ANSWER:

#x=-1/2#
#x=1/4#
#x=-9#

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