Given:
[tex]3u^2-24=0\text{ and }4r^2+432=0.[/tex]Aim:
We need to find the values of u and r.
Explanation:
[tex]Consider\text{ }3u^2-24=0.[/tex]Divide both sides of the equation by 3.
[tex]\text{ }\frac{3u^2}{3}-\frac{24}{3}=\frac{0}{3}.[/tex][tex]u^2-8=0.[/tex]Add 8 to thte both sides of the equation.
[tex]u^2-8+8=0+8.[/tex][tex]u^2=8[/tex]Take sqaure root on both sides.
[tex]\sqrt{u^2}=\pm\sqrt{8}[/tex][tex]Use\text{ }\sqrt{u^2}=u\text{.}[/tex][tex]u=\pm\sqrt{8}[/tex][tex]u=\pm\sqrt{2\times4}[/tex][tex]u=\pm\sqrt{2\times2^2}[/tex][tex]u=\pm2\sqrt{2}[/tex][tex]u=-2\sqrt{2},2\sqrt{2}[/tex][tex]Consider\text{ }4r^2+432=0.[/tex]Divide both sides of the equation by 4.
[tex]\frac{4r^2}{4}+\frac{432}{4}=\frac{0}{4}.[/tex][tex]r^2+108=0[/tex]Subtract 108 from both sides of the equation.
[tex]r^2+108-108=0-108[/tex][tex]r^2=-108[/tex]Take square root on both sides of the equation.
[tex]\sqrt{r^2}=\pm\sqrt{-108}[/tex][tex]\sqrt{r^2}=\pm\sqrt{(-1)\times108}[/tex][tex]We\text{ know that there is no real solution for }\sqrt{(-1)}.[/tex]Final answer:
[tex]u=-2\sqrt{2},2\sqrt{2}[/tex][tex]r=DNE[/tex]