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

Question 1136685: : If one root of the equation x^2-px+20=0 is four, while the equation x^2-qx+p=0 has equal roots then, then a possible value of q is
3
4
5
6
Answer by ikleyn(52778) (Show Source): You can put this solution on YOUR website!
.

The condition says that  x^2 - px + 20 = 0  has the root 4.


Then the other root is   = 5,  according to Vieta's theorem.


It implies (via Vieta's theorem, again) that  p  is equal to the sum of the roots  5+4 = 9:  p = 9.


So, the second equation is


    x^2 - qx + 9 = 0.


Since it has equal roots,  these roots are EITHER


    a)  both are equal to   =  3   (Vieta;s theorem).

OR

    b)  both are equal to   = -3   (Vieta;s theorem).


In case a),  q= 3 + 3 = 6.   (Vieta;s theorem, again)    ANSWER

In case b),  q = (-3) + (-3) = -6.


So, the problem has TWO answers:  q = 6  and/or  q = -6.

Again : one possible answer is q = 6; the other possible answer is q = -6.

Solved.

This problem is to apply Vieta's theorem 4 times.

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