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kearsten

  • one year ago

fan and medal include steps plz 0=8x^2 number of real number solutions

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  1. campbell_st
    • one year ago
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    the only solution is x = 0

  2. kearsten
    • one year ago
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    how did you get that?

  3. campbell_st
    • one year ago
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    well what happnens if you replace x with 0 the equation is \[8 \times 0^2 \] and multiplying anything by 0 is 0 the other parts.... 8 can't be 0..... and the answer when squaring something is dependant on what you are squaring...

  4. kearsten
    • one year ago
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    ok thnx

  5. kearsten
    • one year ago
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    but that not what i needed

  6. campbell_st
    • one year ago
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    what do you need..?

  7. campbell_st
    • one year ago
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    0 is a real number

  8. kearsten
    • one year ago
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    i needed the number of solutions

  9. campbell_st
    • one year ago
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    1 solution... x= 0

  10. campbell_st
    • one year ago
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    well in actual fact there are 2 solutions.... both zero since its a degree 2 polynomial the degree or power of the leading term determines the number of solutions... the can be real or imaginary in your question they are both real

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