Given:
[tex]\begin{gathered} t^2+1t-30=0 \\ t=\frac{N\pm\sqrt[]{D}}{M} \end{gathered}[/tex]Find: N,D,M and value of t.
Sol:.
Quadratic formula:
[tex]\begin{gathered} ax^2+bx+c=0 \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}[/tex][tex]\begin{gathered} t^2+1t-30=0 \\ a=1 \\ b=1 \\ c=-30 \end{gathered}[/tex][tex]\begin{gathered} t=\frac{-1\pm\sqrt[]{1^2-4(1)(-30)}}{2(1)} \\ t=\frac{-1\pm\sqrt[]{1+120}}{2} \\ t=\frac{-1\pm\sqrt[]{121}}{2} \end{gathered}[/tex]So
[tex]\begin{gathered} \frac{-1\pm\sqrt[]{121}}{2}=\frac{N\pm\sqrt[]{D}}{M} \\ N=-1 \\ D=121 \\ M=2 \end{gathered}[/tex](B)
Value of "t"
[tex]\begin{gathered} t=\frac{-1\pm\sqrt[]{121}}{2} \\ t=\frac{-1\pm11}{2} \\ t=\frac{-1+11}{2},t=\frac{-1-11}{2} \\ t=\frac{10}{2},t=\frac{-12}{2} \\ t=5,-6 \end{gathered}[/tex]So value of t is 5,-6
