If the ratio of the roots of the quadratic equation x^2−28x+A=0 is 2:5, what is the product of the roots of the quadratic equation x^2+Ax+2A+3=0?
323
(x-2)(x-5) = x^2-7x+10
we have 28, not 7, so scale up by 4:
(x-8)(x-20) = x^2-28x+160
That makes our cubic
x^2 + 160x + 323 = 0
To find the product of the roots of the quadratic equation x^2+Ax+2A+3=0, we need to determine the roots first.
Let's start by finding the roots of the equation x^2−28x+A=0. We are given that the ratio of the roots is 2:5.
The sum of the roots of a quadratic equation of the form ax^2+bx+c=0 is given by -b/a, and the product of the roots is given by c/a.
In this case, the sum of the roots of x^2−28x+A=0 is 28, and the first root (let's call it r1) is 2x and the second root (let's call it r2) is 5x, where x is a constant.
We know that r1 + r2 = 28, so substituting the values, we get:
2x + 5x = 28
7x = 28
x = 4
Therefore, the roots of the equation x^2−28x+A=0 are 8 and 20.
Now, let's substitute these values into the quadratic equation x^2+Ax+2A+3=0 to find the product of the roots.
Substituting the first root (8) and the second root (20) into the equation, we get:
(8)^2 + A(8) + 2A + 3 = 0
64 + 8A + 2A + 3 = 0
66 + 10A = 0
10A = -66
A = -6.6
Therefore, the product of the roots of the quadratic equation x^2+Ax+2A+3=0 is given by the product of the roots of x^2−28x+A=0, which is 8 * 20 = 160.
So, the product of the roots of x^2+Ax+2A+3=0 is 160.