How do you solve 3+sqrt[x+7]=sqrt[x+4] and find any extraneous solutions?

2 Answers
Gerardina C.
Jun 16, 2016

the equation is impossible

Explanation:

you can calculate

(3+sqrt(x+7))^2=(sqrt(x+4))^2

9+x+7+6sqrt(x+7)=x+4

that's

6sqrt(x+7)=cancel(x)+4-9cancel(-x)-7

6sqrt(x+7)=-12

that's impossible because a square root must be positive

Mia
Jun 16, 2016

No real roots of x exist in R (x!inR)

x is a complex number x=4*i^4-7

Explanation:

First to solve this equation we think how to take off the square root, by squaring both sides:

(3+sqrt(x+7))^2=(sqrt(x+4))^2

Using the binomial property for squaring of sum
(a+b)^2=a^2+2ab+b^2

Applying it on both sides of the equation we have:

(3^2+2*3*sqrt(x+7)+(sqrt(x+7))^2)=x+4

Knowing that (sqrt(a))^2=a
9+6sqrt(x+7)+x+7=x+4

Taking all the know a and unknowns to the second side leaving the square root on one side we have:

6sqrt(x+7)=x+4-x-7-9
6sqrt(x+7)=-12
sqrt(x+7)=-12/6
sqrt(x+7)=-2

Since square root equal to a negative real number that is
impossible in R, no roots exists so we have to check complex set.

sqrt(x+7)=-2

Knowing that i^2=-1 that means -2=2*i^2

sqrt(x+7)=2i^2
Squaring both sides we have:

x+7=4*i^4
Therefore , x=4*i^4-7

So x is a complex number.

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