# How do you solve \sqrt { 169- x ^ { 2} } + 17= x?

Jul 3, 2018

Null set

#### Explanation:

$\sqrt{169 - {x}^{2}} + 17 = x$

$\sqrt{169 - {x}^{2}} = x - 17$

$169 - {x}^{2} = {\left(x - 17\right)}^{2}$

$169 - {x}^{2} = {x}^{2} - 34 x + 289$

$2 {x}^{2} + 34 x - 120 = 0$

${x}^{2} + 17 x - 60 = 0$

$\left(x + 20\right) \cdot \left(x - 3\right) = 0$

But, neither $x = - 20$ nor $x = 3$ are solutions of this equation. Hence, solution of it is null set.

Jul 3, 2018

$\textcolor{b l u e}{\text{No solutions for } x \in \mathbb{R}}$

#### Explanation:

$\sqrt{169 - {x}^{2}} + 17 = x$

Subtract 17 from both sides:

$\sqrt{169 - {x}^{2}} = x - 17$

Square both sides:

$169 - {x}^{2} = {\left(x - 17\right)}^{2}$

Expand RHS:

$169 - {x}^{2} = {x}^{2} - 34 x + 289$

Simplify:

$2 {x}^{2} - 34 x + 120 = 0$

Factor:

$\left(2 x - 10\right) \left(x - 12\right) = 0 \implies x = 5 \mathmr{and} x = 12$

Checking solutions:

$x = 5$

$\sqrt{169 - {\left(5\right)}^{2}} + 17 = 5$

$12 + 17 = 5 \textcolor{w h i t e}{88}$ False

$- 12 + 17 = 5 \textcolor{w h i t e}{88}$ True

$x = 12$

$\sqrt{169 - {\left(12\right)}^{2}} + 17 = 12$

$5 + 17 = 12 \textcolor{w h i t e}{88}$ False

$- 5 + 17 + 12 \textcolor{w h i t e}{88}$True

We can see that the roots found are only solutions if we take the negative root of $\sqrt{169 - {x}^{2}}$, since we assume $\sqrt{169 - {x}^{2}}$ to be the positive root no solutions exist:

Jul 3, 2018

Solution: $x = 12 , x = 5$

#### Explanation:

$\sqrt{169 - {x}^{2}} + 17 = x$ or

$x - 17 = \sqrt{169 - {x}^{2}}$ squaring both sides we get,

${\left(x - 17\right)}^{2} = 169 - {x}^{2}$ or

${x}^{2} - 34 x + 289 = 169 - {x}^{2}$ or

$2 {x}^{2} - 34 x + 289 - 169 = 0$ or

$2 {x}^{2} - 34 x + 120 = 0$ or

$2 \left({x}^{2} - 17 x + 60\right) = 0$ or

${x}^{2} - 17 x + 60 = 0$ or

${x}^{2} - 12 x - 5 x + 60 = 0$ or

$x \left(x - 12\right) - 5 \left(x - 12\right) = 0$ or

$\left(x - 12\right) \left(x - 5\right) = 0 \therefore x = 12 , x = 5$

Solution: $x = 12 , x = 5$ [Ans]