If a b c are non zero,unequal rational numbers then prove that the roots of the eqn (abc²) x²+ 3a²cx+b²cx-6a²-ab+2b²=0 are rational.
To prove that the roots of the given equation are rational, we need to show that they can be expressed as a fraction of two integers.
Let's start by simplifying the equation:
(abc²) x² + 3a²cx + b²cx - 6a² - ab + 2b² = 0
Now, we can apply the quadratic formula to find the roots of the equation:
x = (-b ± √(b² - 4ac)) / (2a)
In this case, the coefficients are:
a = abc²
b = 3a²c + b²c
c = -6a² - ab + 2b²
Substituting these values into the quadratic formula, we get:
x = (-(3a²c + b²c) ± √((3a²c + b²c)² - 4abc²(-6a² - ab + 2b²))) / (2abc²)
To prove that the roots are rational, we need to show that both the numerator and the denominator can be expressed as a fraction of two integers.
Let's simplify the expression step-by-step:
Numerator:
N = -(3a²c + b²c) ± √((3a²c + b²c)² - 4abc²(-6a² - ab + 2b²))
= -3ac(a + b) ± √(((3ac(a + b))^2) + 8abc²(3a + b))
Denominator:
D = 2abc²
As we can see, the numerator contains terms such as (a + b), (3ac(a + b))^2, (3a + b), and constant terms like 8abc².
Since a, b, and c are rational numbers, addition, subtraction, and multiplication of rational numbers also results in a rational number. Therefore, the numerator N is rational.
The denominator D contains only rational numbers (abc²), so it is also rational.
Thus, both the numerator and the denominator are rational, which means the roots of the given equation are rational numbers.