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Math Help - find the range of values for p inequality question

  1. #1
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    Exclamation find the range of values for p inequality question

    Find the range of values of p for which the equation x^2 - 6x + p^2 =0 has real roots

    I got till this part :
    p^2 <= 9
    The answer is -3<=p<=3
    How do I get that? THANK YOU!
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  2. #2
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    Re: find the range of values for p inequality question

    Quote Originally Posted by pumbaa213 View Post
    Find the range of values of p for which the equation x^2 - 6x + p^2 =0 has real roots
    The answer is -3<=p<=3 How do I get that? THANK YOU!
    Use the discriminate: 36-4p^2>0 .
    Thanks from topsquark and pumbaa213
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    Re: find the range of values for p inequality question

    Sorry im still unsure can you help me further? Thank you!
    Why can't The answer be p<=3 OR p<= -3 ? How does it become -3<= p <= 3 ? Thank you!
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    Re: find the range of values for p inequality question

    Quote Originally Posted by pumbaa213 View Post
    Sorry im still unsure can you help me further? How does it become -3<= p <= 3 ? Thank you!
    It is basic algebra. You need to be able to do simple elementary algebra in order to work at this level.

    Maybe you ought to speak to your instructor?
    Thanks from pumbaa213
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    Re: find the range of values for p inequality question

    Quote Originally Posted by pumbaa213 View Post
    Sorry im still unsure can you help me further? Thank you!
    Why can't The answer be p<=3 OR p<= -3 ? How does it become -3<= p <= 3 ? Thank you!
    Think of the number 0 for example. This is \displaystyle \begin{align*} \leq 3 \end{align*}, but it's not \displaystyle \begin{align*} \leq -3 \end{align*}. How are you supposed to know which is correct?
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    Re: find the range of values for p inequality question

    p^2<=9 so p<=sqrt 9 or next part we change the sign wen we take negatve sqrt 9 so p<=3 or p>=-3


    -3<=p<=3
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    Re: find the range of values for p inequality question

    Ok I ve got it..! Thanks guys! :-)
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