if a=5 b=108 c=190
then find the equation whose roots are double the root s of a,b,c......
(x-10)(x-216)(x-380) = 0
You can multiply that out to get a messy cubic equation.
The roots will be 10, 108 and 190, which are twice a, b and c.
Is that what the question asks for? It is a bit confusing. "The roots of a,b,c.." doesn't make much sense.
yeh man the roots of abc
a, b and c are constants and have no roots. Equations have roots
To find the equation whose roots are double the roots of a, b, and c, you can start by determining the quadratic equation using the given values of a, b, and c.
Let's consider the quadratic equation:
ax^2 + bx + c = 0
Given that a = 5, b = 108, and c = 190, we can substitute these values into the equation:
5x^2 + 108x + 190 = 0
Now, we need to find the equation whose roots are double the roots of the given equation. Let's call the roots of the given equation as r1 and r2. The new equation will have roots 2r1 and 2r2.
So, the new equation can be written as:
5(x - 2r1)(x - 2r2) = 0
Simplifying this equation, we get:
5(x^2 - (2r1 + 2r2)x + 4r1r2) = 0
To relate this equation to the given equation, we can use Vieta's formulas, which state that in a quadratic equation ax^2 + bx + c = 0, the sum of the roots is -b/a, and the product of the roots is c/a.
For the given equation:
Sum of the roots = -(b/a) = -(108/5) = -21.6
Product of the roots = c/a = 190/5 = 38
Since the roots of the new equation are double the roots of the given equation, we can use these formulas to calculate the roots of the new equation:
Sum of new roots = 2 times the sum of the roots = 2 * (-21.6) = -43.2
Product of new roots = 2 times the product of the roots = 2 * 38 = 76
Hence, the equation whose roots are double the roots of the given equation is:
5(x^2 + 43.2x + 76) = 0