How do you solve #|2x+5| = 3x+4# and find any extraneous solutions?

2 Answers
Apr 4, 2017

Answer:

Non-extraneous solution #x=1#

Explanation:

Given
#color(white)("XXX")abs(2x+5)=3x+4#

One method of solving this would be to square both sides:
#color(white)("XXX")4x^2+20x+25=9x^2+24x+16#

Simplifying
#color(white)("XXX")5x^2+4x-9=0#

Factoring
#color(white)("XXX")(5x+9)(x-1)=0#

Giving potential solutions
#color(white)("XXX"){:(x=-9/5,color(white)("XX")andcolor(white)("XX"),x=+1):}#

Checking these against the original equation:
#color(white)("XXX"){: (ul(color(white)("XXXXXX")),ul(x=-9/5),color(white)("XXX"),ul(x=1)), (abs(2x+5),=abs(-18/5+25/5),,=abs(2+5)), (,=7/5,,=7), (ul(color(white)("XXXXXX")),ul(color(white)("XXXXXX")),,ul(color(white)("XXXXXX"))), (3x+4,=-27/5+20/5,,=3+4), (,=-7/5,,=7) :}#

Since #abs(2x+5)!=3x+4# when #x=-9/5#
this solution is extraneous.

Apr 4, 2017

Answer:

#x=1," extraneous solution is " x=-9/5#

Explanation:

This equation has 2 solutions.

#rArr2x+5=color(red)(+-)3x+4#

#• " solve " 2x+5=3x+4#

subtract 2x from both sides.

#cancel(2x)cancel(-2x)+5=3x-2x+4#

#rArr5=x+4#

subtract 4 from both sides.

#5-4=xcancel(+4)cancel(-4)#

#rArrx=1larrcolor(red)" to be verified"#

#• " solve " 2x+5=color(red)(-)(3x+4)#

#rArr2x+5=-3x-4#

add 3x to both sides.

#2x+3x+5=cancel(-3x)cancel(+3x)-4#

#rArr5x+5=-4#

subtract 5 from both sides.

#5xcancel(+5)cancel(-5)=-4-5#

#rArr5x=-9#

divide both sides by 5

#(cancel(5) x)/cancel(5)=(-9)/5#

#rArrx=-9/5larrcolor(red)" to be verified"#

#color(blue)"As a check"#

Substitute these values into the equation and if the left side equals the right side then they are the solutions.

#x=1to|2+5|=3+4to7=7to" true"#

#x=-9/5to|-18/5+25/5|=-27/5+20/5to7/5=-7/5to" false"#

#rArrx=1" is the solution"#

#rArrx=-9/5" is an extraneous solution"#