Question

How do you solve `sqrt(7x+109) = 2x+17`?

Answer 1

`x= -4`

Explanation:

Square both sides.

`sqrt(7x+109)^2 = (2x+17)^2`

`7x + 109 = 4x^2 + 68x + 289`

Now subtract `7x + 109` from both sides.

`0 = 4x^2 + 61x + 180`

This gives us a quadratic where `a=4, b=61, c=180`

Use the quadratic formula:

`x = (-61 +- sqrt(61^2 - 4(4)(180)))/(2*4)`

`x = (-61 +- sqrt(3721 - 2880))/8`

`x = (-61 +- sqrt(841))/8`

`x = (-61+-29)/8`

`x = (-90)/8 " " or " " x = (-32)/8`

`x = (-45)/4 " " or " " x = -4`

Since we squared both sides, there's a chance that one or both of our solutions are extraneous (appear correct but aren't when you plug them back in). So, let's plug both solutions back in and see if they work:

`sqrt(7(-4)+109) stackrel(?color(white)"x")(=) 2(-4)+17`

`sqrt(109 - 28) stackrel(?color(white)"x")(=) 17 - 8`

`sqrt81 stackrel(?color(white)"x")(=) 9`

`9 stackrel(color(limegreen)sqrt()color(white)"X.")(=) 9`

This one works! So `x = -4` is a valid solution. Now for the other one:

`sqrt(7(-45/4)+109) stackrel(?color(white)"x")(=) 2(-45/4)+17`

`sqrt(-315/4 + 436/4) stackrel(?color(white)"x")(=) -45/2 + 34/2`

`sqrt(121/4) stackrel(?color(white)"x")(=) -11/2`

`11/2 ne -11/2`

So this solution is extraneous. Note that the only reason we might have thought it was true is because we squared both sides `-` although

`11/2 = -11/2`

is a false statement, the statement

`(11/2)^2 = (-11/2)^2`

is true. This is why it's always important to check for extraneous solutions when working with quadratics like this.

So our solution is `x = -4`

Final Answer