Question

For each of the following equations, say which one (d) has two rational solutions, (e) has two irrational
solutions, and (f) has two non-real solutions. Each answer (d, e, and f) is used once.

Answer 1

to solve this equation

$x { }^{2} + 3x - 40 = 0$

1. we first identify it as quadratic and to solve we find two numbers that can multiply to get -40 and add to get positive three
2. $(x - 5)(x + 8)$
3. the factors being-5 and positive 8 satisfied the condition now to apply zero product principle
4. $(x - 5) = 0 \: and \: (x + 8) = 0$
5. from the equation above in (x-5)=0 we subtract 0 from both places and it literally reads
6. $x = + 5$
7. for the second the same is done
8. $x = - 8$
9. now for second question
10. $x^{2} + 3x + 1$
11. not forgetting the zero. open two brackets and find no that can add to get 3 and multiply to get 1
12. $(x + 1)(x + 1)$
13. the answer being x=-1 and x=-1