Question

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

Answer 1

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.

Answer 2

`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"`