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

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Answer 1 · 2016-08-19T13:47:31

The soln. is #x=14#, and, #x=3# is extraneous soln..

Rewriting the eqn. as, #x-7=sqrt(3x+7)#, and, squaring, we get,

#x^2-14x+49=3x+7#.

#rArr x^2-17x+42=0#.

Using, #14xx3=42, and, 14+3=17#, we have,

#ul(x^2-14x)-ul(3x+42)=0#.

#rArr x(x-14)-3(x-14)=0#.

#rArr (x-14)(x-3)=0#.

#rArr x=14, or, x=3#.

Recall that #sqrt(3x+7)# is always #+ve#, so that, #x=3# does not satisfy the original eqn., &, hence is an extraneous soln.

Thus, the soln. is #x=14#, and, #x=3# is extraneous.