How do you solve #sqrt(2x+7)=x+3#?

1 Answer
Jul 2, 2016

Answer:

Do a little squaring and quadratic-equation solving to get #x=-2+sqrt2#.

Explanation:

The first thing you want to do in radical equations is get the radical on one side of the equation. Today is our lucky day, because that has already been done for us.

Next step is to square both sides to get rid of the radical:
#sqrt(2x+7)=x+3#
#(sqrt(2x+7))^2=(x+3)^2#
#->2x+7=x^2+6x+9#

Now we have to combine like terms and set the equation equal to #0#:
#2x+7=x^2+6x+9#
#0=x^2+(6x-2x)+(9-7)#
#->0=x^2+4x+2#

Unfortunately, this quadratic equation does not factor, so we'll have to use the quadratic formula:
#x=(-b+-sqrt(b^2-4ac))/(2a)#

With #a=1#, #b=4#, and #c=2#, our solutions are:
#x=(-(4)+-sqrt((4)^2-4(1)(2)))/(2(1))#
#x=(-4+-sqrt(16-8))/2#
#x=-4/2+-sqrt(8)/2#
#->x=-2+-sqrt(2)#
(Note that #sqrt(8)/2=(2sqrt(2))/2=sqrt2#)

We have our solutions: #x=-2+sqrt2~~-0.586# and #x=-2-sqrt2~~-3.414#. But because this is an equation involving radicals, we need to double-check our solutions.

Solution 1: #x~~-0.586#
#sqrt(2x+7)=x+3#
#sqrt(2(-0.586)+7)=-0.586+3#
#2.414=2.414-># Solution checks

Solution 2: #x~~-3.414#
#sqrt(2x+7)=x+3#
#sqrt(2(-3.414)+7)=-3.414+3#
#.415!=-.414-># Extraneous solution

As you can see, only one of our solutions work: #x=-2+sqrt2#.