If the roots of the eqn x²-2cx+ab=0 are real and unequal, then prove that the roots of x²-2(a+b)x+a²+b²+2c²=0 will be imaginary.
To prove that the roots of the equation x² - 2(a+b)x + a² + b² + 2c² = 0 will be imaginary, we can use the discriminant.
Let's start by finding the discriminant of the given equation: x² - 2(a+b)x + a² + b² + 2c² = 0.
The discriminant (D) of a quadratic equation ax² + bx + c = 0 is given by D = b² - 4ac.
Comparing the given equation with the general form ax² + bx + c = 0, we have:
a = 1
b = -2(a+b)
c = a² + b² + 2c²
Now, substituting the values of a, b, and c into the formula for the discriminant, we get:
D = (- 2(a+b))² - 4(1)(a² + b² + 2c²)
= 4(a+b)² - 4(a² + b² + 2c²)
= 4(a² + 2ab + b²) - 4a² - 4b² - 8c²
= 4a² + 8ab + 4b² - 4a² - 4b² - 8c²
= 8ab - 8c²
To prove that the roots of the equation x² - 2(a+b)x + a² + b² + 2c² = 0 will be imaginary, we need to show that the discriminant (D) is negative, i.e., D < 0.
Substituting 8ab - 8c² into D < 0, we have:
8ab - 8c² < 0
ab - c² < 0
Since the roots of the original equation x² - 2cx + ab = 0 are real and unequal, we know that the discriminant of that equation which is ab - 4(ab) = -3ab < 0.
Therefore, ab - c² < 0 and -3ab < 0, imply that:
ab - c² < -3ab
c² > 4ab
This shows that the discriminant of the new equation x² - 2(a+b)x + a² + b² + 2c² = 0 is negative. Hence, the roots of the equation will be imaginary.