Question

Show that the equation 2(a²+b²)x²+2(a+b)x+1=0 where a≠b

Answer 1

Answer:

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

Answer 2

plz refer to this attachment