Question

How do you evaluate `(8-t)^2-(t+5-t^2)^2+ (t^2 + 1)^2 -2t (t^2+ 6t -13) > 21`?

Answer 1

First, let's expand and simplify like terms on the left side of the equation.

`(8-t)^2-(t+5-t^2)^2+ (t^2 + 1)^2 -2t (t^2+ 6t -13) > 21 ->`

`(t^2-16t+64)+(-t^4+2t^3+9t^2-10t-25)+(t^4+t^2+1)+(-2t^3-12t^2+26t) ->`

`(-t^4+t^4)+(2t^3-2t^3)+(t^2+9t^2+t^2-12t^2)+(-16t-10t+26t)+(64-25+1) ->`

`(0t^4)+(0t^3)+(0t^2)+(0t)+(40) ->`

`40`

Let's plug this simplified version back into the inequality.

`40 > 21`

This is a true statement. When you get a true statement, that means that t is all numbers or `t ∈ -oo≤t≤oo`.