Question

How do you solve `sqrt (3x + 4) + x = 8`?

Answer 1

`color(blue)(x=16)`

`color(blue)(x=4)`

Explanation:

`sqrt(3x+4)+x=8`

Subtract `x` from both sides:

`sqrt(3x+4)=8-x`

Square both sides:

`3x+4=(8-x)^2`

Expand the bracket:

`3x+4=64-16x+x^2`

Collect like terms:

`x^2-19x+60=0`

Factor:

`x^2-15x-4x+60`

`x(x-15)-4(x-15)`

`(x-15){x-4}`

`(x-16)(x-4)=0`

`x-16=0=>color(blue)(x=16)`

`x-4=0=>color(blue)(x=4)`

Answer 2

Solution: `x=4 , x=15`

Explanation:

`sqrt(3x+4)+x =8 or sqrt(3x+4) =8-x` Squaring both sides

we get, `3x+4 = (8-x)^2 or 3x+4 = x^2-16x+64` or

`x^2-16x+64-3x-4=0 or x^2-19x+60=0` or

`x^2-15x-4x+60=0` or

`x(x-15)-4(x-15)=0 or (x-15)(x-4)=0 :.`

Either `x-15=0 :. x=15` , or `x-4=0 :. x=4`

Check : `sqrt(3*4+4)+4 =8 or sqrt16+4=8or 8=8`

`sqrt(3*15+4)+15 =8 or sqrt49+15=8 or -7+15=8`

Solution: `x=4 , x=15` [Ans]