Question

Show that the equation 2(a²+b²)x²+2(a+b)x+1=0 has no real roots, when a≠b .

Answer 1

The roots of the equation 2 (a²+b²) x² + 2 (a+b) x + 1 = 0 are not real as the discriminant is equal to - 2 (3a² + 3b² - 2ab).

• A quadratic equation has no real roots when its discriminant is less than zero.

• A discriminant is given by the formula b² - 4ac. Therefore, for the roots to be imaginary, the following condition should be followed :

b² - 4ac < 0

• Now, for the given quadratic equation,

a = coefficient of x² = 2 (a² + b²)

b = coefficient of x = 2 (a + b)

c = constant = 1

• b² - 4ac = 2 (a + b)² - 4.2 (a² + b²).1

= 2 (a² + b² + 2ab) - 8 (a² + b²)

= 2a² + 2b² + 4ab - 8a² - 8b²

= 2a² - 8a² + 2b² - 8b² + 4ab

= - 6a² - 6b² + 4ab

= - 2 (3a² + 3b² - 2ab)

• Since the discriminant b² - 4ac is a negative value ( < 0),  it implies that the roots of the given quadratic equation are not real.

Answer 2

plz refer to this attachment