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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.

For each of the following equations say which one d has two rational solutions e has two irrational solutions and f has two nonreal solutions Each answer d e an class=

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paulnneji

to solve this equation

[tex]x { }^{2} + 3x - 40 = 0[/tex]

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

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