What is the value of #x# in the equation #sqrtx + 5 = sqrt(x+45)#?

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Answer 1 · 2016-04-09T22:36:49

#x=4#

In this case you have to square the sides od the equation because:

#color(blue)(a=b=>a^2=b^2#

But also
#color(red)(a=-b=>a^2=b^2#, so we can add false solutions when trying to find the solutions.

#sqrt(x)+5=sqrt(x+45)#

Squaring;

#x +10sqrt(x)+25=x+45#

#10sqrt(x)=20#

#sqrt(x)=2#

Squaring again:

#x=4#

If we try it we realise that x=4 is an actual solution.