SOLUTION: If one root of the equation x^2-px+20=0 is four, while the equation x^2-qx+20=0 has equal roots then, then a possible value of q is 3 4 5 6

Algebra ->  Numeric Fractions Calculators, Lesson and Practice -> SOLUTION: If one root of the equation x^2-px+20=0 is four, while the equation x^2-qx+20=0 has equal roots then, then a possible value of q is 3 4 5 6      Log On


   



Question 1136680: If one root of the equation x^2-px+20=0 is four, while the equation x^2-qx+20=0 has equal roots then, then a possible value of q is
3
4
5
6

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
If one root of the equation x^2-px+20=0 is four, while the equation x^2-qx+20=0 has equal roots then, then a possible value of q is
3
4
5
6
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Regarding this post, I'd like to make couple of notices.

    1.  First statement "one root of the equation x^2-px+20=0 is four" has no any relation and any connection with the rest of the post.


    2.  The valid value of " q " is not in the option list.


So,  based on these notices,  I conclude that the problem in the post is,  strictly saying,   D E F E C T I V E.


The answer to the problem question is  q = 2%2Asqrt%2820%29.


But it is not relevant to the first cited statement.