Solve for all possible values of x. sqrt (19-5x)=x-5 Answer Attemptiout of 3 (4) Additional Solution No Solution x=square

To solve the equation $\sqrt{19-5x} = x-5$, we need to find the values of $x$ that satisfy the equation.<br /><br />Step 1: Square both sides of the equation to eliminate the square root.<br />$(\sqrt{19-5x})^2 = (x-5)^2$<br />$19-5x = x^2 - 10x + 25$<br />$0 = x^2 - 15x + 6$<br /><br />Step 2: Solve the quadratic equation $x^2 - 15x + 6 = 0$.<br />We can use the quadratic formula to solve this equation:<br />$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$<br />where $a = 1$, $b = -15$, and $c = 6$.<br /><br />$x = \frac{15 \pm \sqrt{225 - 24}}{2}$<br />$x = \frac{15 \pm \sqrt{201}}{2}$<br /><br />Step 3: Check the solutions.<br />The solutions are $x = \frac{15 + \sqrt{201}}{2}$ and $x = \frac{15 - \sqrt{201}}{2}$.<br /><br />However, we need to check if these solutions satisfy the original equation $\sqrt{19-5x} = x-5$.<br /><br />For $x = \frac{15 + \sqrt{201}}{2}$:<br />$\sqrt{19-5(\frac{15 + \sqrt{201}}{2})} = \frac{15 + \sqrt{201}}{2} - 5$<br />$\sqrt{19-5(\frac{15 + \sqrt{201}}{2})} = \frac{15 + \sqrt{201}}{2} - 5$<br />$\sqrt{19-5(\frac{15 + \sqrt{201}}{2})} = \frac{15 + \sqrt{201}}{2} - 5$<br />$\sqrt{19-5(\frac{15 + \sqrt{201}}{2})} = \frac{15 + \sqrt{201}}{2}5$<br />$\sqrt{19-5(\frac{15 + \sqrt{201}}{2})} = \frac{15 + \sqrt{201}}{2} - 5$<br />$\sqrt{19-5(\frac{15 + \sqrt{201}}{2})} = \frac{15 + \sqrt{201}}{2} - 5$<br />$\sqrt{19-5(\frac{15 + \sqrt