Respuesta :
[tex]\large\displaystyle\text{$\begin{gathered}\sf \sqrt{z^{2}-8 }-\sqrt{6z-17}=0 \end{gathered}$}[/tex]
Simplify the left side. Yes n is a positive integer greater than x and a is a real number or a factor, then [tex]\bf{\sqrt[n]{a^{x} }=a^{\frac{x}{n} }. }[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{(z^{2}-8)^{\frac{1}{2} }-\sqrt{6z-17}=0 } \end{gathered}$}[/tex]
If n is a positive integer greater than x and a is a real number or a factor, then [tex]\bf{\sqrt[n]{a^{x} }=a^{\frac{x}{n} }. }[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf \bf{(z^{2}-8)^{\frac{1}{2} }-(6z-17)^{\frac{1}{2}} =0 } \end{gathered}$}[/tex]
Draw each side of the equation. The solution is the x value of the point of intersection.
[tex]\red{\underbrace{\overbrace{\boxed{\boldsymbol{\sf{\green{z=3}}}}} \ \ \to \ \ \ Answer}}[/tex]
{ Pisces04 }
Answer:
Step-by-step explanation:
[tex]\sqrt{z^2-8} -\sqrt{6z-17} =0\\\sqrt{z^2-8} =\sqrt{6z-17} \\6z > 17\\z > \frac{17}{6} \\z^2-8=0\\z^2 > 8\\so \\z < -\sqrt{8} \\or\\z > \sqrt{8} \\combining\\z > \frac{17}{6} \\again\\z^2-8=6z-17\\z^2-6z-8+17=0\\z^2-6z+9=0\\z^2-3z-3z+9=0\\z(z-3)-3(z-3)=0\\(z-3)(z-3)=0\\z=3[/tex]
