The equation x^2+px+q=0, q cannot be equal to 0, has two unequal roots such that the squares of the roots are the same as the two roots. Calculate the product pq.
I think the obvious root would be one but the second roots i just can't figure out!
a web search will turn up this exact problem and its solution
To find the roots of the quadratic equation x^2 + px + q = 0, we can use the quadratic formula:
x = (-p ± √(p^2 - 4q)) / 2
Since the squares of the roots are equal to the two roots, we can write it as:
x^2 = x
Simplifying, we get:
x^2 - x = 0
This equation can be factored as:
x(x - 1) = 0
Therefore, the two potential roots are x = 0 and x = 1.
Since the question states that the roots are unequal, we can conclude that there is one root equal to 1 and the other root equal to 0.
Now, let's substitute these roots into the equation x^2 + px + q = 0:
(1)^2 + p(1) + q = 0
This simplifies to:
1 + p + q = 0
Likewise for x = 0:
(0)^2 + p(0) + q = 0
This simplifies to:
q = 0
But the problem states that q cannot be equal to 0, which means this solution is not valid.
Therefore, we have only one solution: x = 1.
Substituting this back into the equation:
(1)^2 + p(1) + q = 0
This simplifies to:
1 + p + q = 0
Since q ≠ 0, we can rearrange the equation to find q in terms of p:
q = -(p + 1)
Now, we can calculate the product pq:
pq = p(-p - 1) = -p^2 - p
Therefore, the product pq is -p^2 - p.