Answer:
-IF THE EQUATION IS [tex]\frac{1}{t-2}=\frac{t}{8}[/tex], THEN:
[tex]t_1=4\\t_2=-2[/tex]
-IF THE EQUATION IS [tex]\frac{1}{t}-2=\frac{t}{8}[/tex], THEN:
[tex]t_1=-16.485\\t_2=0.485[/tex]
Step-by-step explanation:
-IF THE EQUATION IS [tex]\frac{1}{t-2}=\frac{t}{8}[/tex] THE PROCEDURE IS:
Multiply both sides of the equation by [tex]t-2[/tex] and by 8:
[tex](8)(t-2)(\frac{1}{t-2})=(\frac{t}{8})(8)(t-2)\\\\(8)(1)=(t)(t-2)\\\\8=t^2-2t[/tex]
Subtract 8 from both sides of the equation:
[tex]8-8=t^2-2t-8\\\\0=t^2-2t-8[/tex]
Factor the equation. Find two numbers whose sum be -2 and whose product be -8. These are -4 and 2:
[tex]0=(t-4)(t+2)[/tex]
Then:
[tex]t_1=4\\t_2=-2[/tex]
-IF THE EQUATION IS [tex]\frac{1}{t}-2=\frac{t}{8}[/tex] THE PROCEDURE IS:
Subtract [tex]\frac{1}{t}[/tex] and [tex]2[/tex]:
[tex]\frac{1}{t}-2=\frac{t}{8}\\\\\frac{1-2t}{t}=\frac{t}{8}[/tex]
Multiply both sides of the equation by [tex]t[/tex]:
[tex](t)(\frac{1-2t}{t})=(\frac{t}{8})(t)\\\\1-2t=\frac{t^2}{8}[/tex]
Multiply both sides of the equation by 8:
[tex](8)(1-2t)=(\frac{t^2}{8})(8)\\\\8-16t=t^2[/tex]
Move the [tex]16t[/tex] and 8 to the other side of the equation and apply the Quadratic formula. Then:
[tex]t^2+16t-8=0[/tex]
[tex]t=\frac{-b\±\sqrt{b^2-4ac}}{2a}\\\\t=\frac{-16\±\sqrt{16^2-4(1)(-8)}}{2(1)}\\\\t_1=-16.485\\t_2=0.485[/tex]
