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 quadric equation is 2(a² + b²)x² + 2(a + b)x + 1 = 0

Here,

a = 2(a² + b²), b = 2(a + b) and c = 1

As we know that D = b² - 4ac

Putting the value of a = 2(a² + b²), b = 2(a + b) and c = 1

D = {2(a + b)}² - 4 x (2(a² + b²)) x (1)

= 4(a² + 2ab + b²) - 8(a² + b²)

= 4a² + 8ab + 4b² - 8a² - 8b²

= 8ab - 4a² - 4b²

= -4(a² - 2ab + b²)

= -4(a - b)²

We have,

a ≠ b

a - b ≠ 0

Thus, the value of D < 0

Therefore, the roots of the given equation are not real

Hence, proved