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

Jun 16, 2016

the equation is impossible

#### Explanation:

you can calculate

${\left(3 + \sqrt{x + 7}\right)}^{2} = {\left(\sqrt{x + 4}\right)}^{2}$

$9 + x + 7 + 6 \sqrt{x + 7} = x + 4$

that's

$6 \sqrt{x + 7} = \cancel{x} + 4 - 9 \cancel{- x} - 7$

$6 \sqrt{x + 7} = - 12$

that's impossible because a square root must be positive

Jun 16, 2016

No real roots of $x$ exist in $R$ ($x \notin R$)

$x$ is a complex number $x = 4 \cdot {i}^{4} - 7$

#### Explanation:

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

${\left(3 + \sqrt{x + 7}\right)}^{2} = {\left(\sqrt{x + 4}\right)}^{2}$

Using the binomial property for squaring of sum
${\left(a + b\right)}^{2} = {a}^{2} + 2 a b + {b}^{2}$

Applying it on both sides of the equation we have:

$\left({3}^{2} + 2 \cdot 3 \cdot \sqrt{x + 7} + {\left(\sqrt{x + 7}\right)}^{2}\right) = x + 4$

Knowing that ${\left(\sqrt{a}\right)}^{2} = a$
$9 + 6 \sqrt{x + 7} + x + 7 = x + 4$

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

$6 \sqrt{x + 7} = x + 4 - x - 7 - 9$
$6 \sqrt{x + 7} = - 12$
$\sqrt{x + 7} = - \frac{12}{6}$
$\sqrt{x + 7} = - 2$

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 $- 2 = 2 \cdot {i}^{2}$

$\sqrt{x + 7} = 2 {i}^{2}$
Squaring both sides we have:

$x + 7 = 4 \cdot {i}^{4}$
Therefore , $x = 4 \cdot {i}^{4} - 7$

So $x$ is a complex number.