# How do you solve sqrt( 9 + x) + sqrt(1+x)=sqrt(x + 16)?

May 9, 2015

First of all the existence conditions:

$9 + x \ge 0 \Rightarrow x \ge - 9$

$1 + x \ge 0 \Rightarrow x \ge - 1$

$x + 16 \ge 0 \Rightarrow x \ge - 16$,

the solution of the system of inequalities is $x \ge - 1$.

Now we can square the two members because the first is positive or zero (sum of two quantities positive or zero), and so it is the second.

${\left(\sqrt{9 + x} + \sqrt{1 + x}\right)}^{2} = {\left(\sqrt{x + 16}\right)}^{2} \Rightarrow$

$9 + x + 2 \sqrt{9 + x} \cdot \sqrt{1 + x} + 1 + x = x + 16 \Rightarrow$

$2 \sqrt{9 + x} \cdot \sqrt{1 + x} = x + 16 - 9 - x - 1 - x \Rightarrow$

$2 \sqrt{9 + x} \cdot \sqrt{1 + x} = 6 - x$.

Now we have another contion to solve, because we have to make sure that the second member is positive or zero, because we want to square the two members another time.

$6 - x \ge 0 \Rightarrow x \le 6$ that, joined with the first one $x \ge - 1$, becomes:

$- 1 \le x \le 6$.

$4 \left(9 + x\right) \left(1 + x\right) = {\left(6 - x\right)}^{2} \Rightarrow$

$4 \left(9 + 9 x + x + {x}^{2}\right) = 36 - 12 x + {x}^{2} \Rightarrow$

$36 + 36 x + 4 x + 4 {x}^{2} - 36 + 12 x - {x}^{2} = 0 \Rightarrow$

$3 {x}^{2} + 28 x = 0 \Rightarrow x \left(3 x + 28\right) = 0 \Rightarrow$

${x}_{1} = 0$ acceptable solution (it is in $- 1 \le x \le 6$)

${x}_{2} = - \frac{28}{3}$ not acceptable solution (it is not in $- 1 \le x \le 6$).