How do you solve #\sqrt { 2x + 41} - 3= x#?

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Answer 1 · 2018-03-14T12:09:36

#x=-8,4#

We have:

#x=sqrt(2x+41)-3#

#x+3=sqrt(2x+41)#

#(x+3)^2=2x+41#

#x^2+6x+9=2x+41#

#x^2+4x-32=0#

We can use many methods to solve this. Let's use the quadratic formula. According to it:

#x=(-b+-sqrt(b^2-4ac))/(2a)#

Here, #a=1, b=4, c=-32#. Inputting:

#x=(-4+-sqrt(4^2-(4)(1)(-32)))/(2*1)#

#x=(-4+-sqrt(16+126))/2#

#x=(-4+-12)/2#

#x=-2+-6#

#x=-8,4#