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.
there is only one number which satisfies "the square of itself is equal to itself": it is 1. (0 is also the other but it is stated that q cannot be equal to 0)
therefore,
(x-1)(x-1) = x^2 + 2x + 1
p = 2
q = 1
hope this helps~ :)
To find the product pq, we need to find the value of p and q.
Given that the equation x^2 + px + q = 0 has two unequal roots such that the squares of the roots are the same as the two roots, we can write the two roots as r and s.
According to the given condition, we have:
r^2 = r
s^2 = s
Now, let's find the value of r and s.
From the equation x^2 + px + q = 0, we know that the sum of the roots is equal to -p, and the product of the roots is equal to q.
Using the sum and product of roots formulas, we have:
r + s = -p
rs = q
Solving these equations together will give us the values of r and s.
From the equation r^2 = r, we can see that r = 0 or r = 1.
Case 1: r = 0
If r = 0, then from the equation rs = q, we have 0s = q. Since q ≠ 0 (as given in the problem), we know that s ≠ 0. Therefore, we can conclude that r = 0 would result in both roots being equal, which contradicts the given condition of two unequal roots.
Case 2: r = 1
If r = 1, then from the equation rs = q, we have s = q. In this case, the roots r and s are distinct and unequal. Also, since r^2 = r, we have 1^2 = 1, which satisfies the given condition.
Now, we have r = 1 and s = q. Substituting these values into r + s = -p, we get 1 + q = -p, which implies q = -p - 1.
Since we have found the values of r = 1 and q = -p - 1, we can now calculate the product pq.
pq = (-p - 1) * p
Expanding the equation gives us:
pq = -p^2 - p
Therefore, the product pq is -p^2 - p.