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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Originally Posted by pumbaa213 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: .
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!
Originally Posted by pumbaa213 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?
Originally Posted by pumbaa213 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 , but it's not . How are you supposed to know which is correct?
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
Ok I ve got it..! Thanks guys! :-)
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