+184 The polynomial p(x) = x^2+ax+b has distinct roots 2a and 2b. Find a+b.
Since 2a and 2b are the roots of p(x), we know that
p(x) = (x - 2a)(x - 2b) = x^2 - 4ax + 4ab
Equating coefficients, we get
a = -4a b = 4ab
Solving for a and b, we find
a = -1 b = -2
Therefore, a+b = -3.
+184 Thank you... but the answer is actually -1/4.. still appreciate the effort!
To find the values of a and b, we can use the relationship between the roots and coefficients of a quadratic equation.
In a quadratic equation of the form p(x) = x^2 + ax + b, the sum of the roots is equal to the negation of the coefficient of x (a), and the product of the roots is equal to the constant term (b).
Given that the distinct roots are 2a and 2b, we have the following relationships:
Sum of the roots: 2a + 2b = -a (1) Product of the roots: (2a)(2b) = b (2)
From equation (1), we can simplify it by moving all terms to one side:
2a + 2b + a = 0 3a + 2b = 0
Now, let's solve equations (1) and (2) simultaneously:
From equation (2), we have (2a)(2b) = b, which can be rewritten as 4ab = b. We can divide both sides by b (assuming b is nonzero) to obtain 4a = 1. Dividing both sides by 4, we get a = 1/4.
Substituting the value of a = 1/4 into equation (1): myhdfs login
3(1/4) + 2b = 0 3/4 + 2b = 0 2b = -3/4 b = -3/8
Therefore, the values of a and b are a = 1/4 and b = -3/8, respectively. To find a + b, we can substitute these values:
a + b = 1/4 + (-3/8) = 2/8 - 3/8 = -1/8.
Hence, the value of a + b is -1/8.
