Since the point lies on the unit circle, it satisfies
[tex]x^2+\left(-\dfrac8{17}\right)^2=x^2+\left(\dfrac8{17}\right)^2=1[/tex]
Solving for [tex]x[/tex] gives two possible solutions:
[tex]x^2=1-\left(\dfrac8{17}\right)^2[/tex]
[tex]x=\pm\sqrt{1-\left(\dfrac8{17}\right)^2}[/tex]
Given that [tex]x[/tex] is in quadrant IV, you know that its value must be positive, so
[tex]x=\sqrt{1-\left(\dfrac8{17}\right)^2}=\sqrt{\dfrac{225}{289}}[/tex]
