Question 1132449: 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 a possible value of q is
A) 3
B 4
C 5
D 6
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13195)   (Show Source):
You can put this solution on YOUR website!
One root of x^2-px+20 is 4, so x^2-px+20 = (x-4)(x-a) which means a = 5, so the other root is 5.
Then x^2-px+20 = (x-4)(x-5) = x^2-9x+20, so p is 9.
You can also solve that part of the problem using Vieta's theorem. The product of the roots is 20, so the second root is 5; then the sum of the roots is p, so p = 9.
In the next part of the problem, we are to find the value of q if x^2-qx+p has equal roots.
We now know that p = 9, so this expression is x^2-qx+9. For it to have equal roots, we need to have x^2-qx+9 = (x-a)^2 = x^2-2ax+a^2, which means a = 3; then q = 2a = 6.
ANSWER: D
Answer by ikleyn(52756)  (Show Source):
You can put this solution on YOUR website! .
If one root of the equation x^2-px+20 = 0 is 4, then the other root is  = 5 (according to the Vieta's theorem).
Then the coefficient "p" in this equation is -p = -(4+5) = -9, or p= 9 (by applying the Vieta's theorem, again).
Now let's look into the second equation.
Since both its roots are equal, their product is equal to the constant term, which is 9.
(Here I applied the Vieta's theorem, again).
Hence, these roots are EITHER 3 (both), OR -3 (both).
Then for q the possible values are EITHER (3+3) = 6 OR ((-3) + (-3)) = -6.
ANSWER. There are TWO possibilities for q: 6 or -6.
Notice that your list of possible answers IS UNCOMPLETED (!), which is the fault of the person, who invented/created this problem.
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Comment from student: Answer is D i.e 6 and it is also mentioned in the list
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My response : As I showed in my solution, "q" may have TWO values, 6 and -6.
The value of "-6" is not shown in the answer list.
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