Question

If a and b are distinct real numbers, show that the quadratic equation
2(a sq. + b sq.)x sq. + 2(a+b)x + 1 = 0 has no real roots.​

Answer 1

Solution:

2(a² + b²)x² + 2(a + b)x + 1 = 0

Here, A = 2(a² + b²), B = 2(a + b) and C = 1

The quadratic equation will have non real roots if the value of discriminant is less than zero.

i.e. To prove B² - 4AC < 0

L.H.S. = [2(a + b)]² - 4[ 2(a² +b²) ] (1)

= 4a² + 4b² + 8ab - 8a² - 8b²

= - 4a² + 8ab - 4b²

= - 4(a² - 2ab + b²)

= - 4 (a - b)²

Since, a and b are real distinct numbers square of their multiple will be positive and if is multiplied by - 4 then it will always form a negative number.

i.e. L.H.S. < 0

i.e. L.H.S. = R.H.S.

Therefore, the given quadratic equation will not have any real roots.