12th q
Q.12. Show that the equation 2 a 2 + b 2 x 2 + 2 a + b x + 1 = 0 has no real roots.

Given quadratic equation is

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

Comparing with Ax2 + Bx + C = 0, we get

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

Now, calculate D = B2 - 4ac

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

=> D = 4(a2 + b2 + 2ab) - 8(a2 + b2 )

=> D = 4a2 + 4b2 + 8ab - 8a2 - 8b2

=> D = -4a2 - 4b2 + 8ab

=> D = -(4a2 + 4b2 - 8ab)

=> D = -4(a2 + b2 - 2ab)

=> D = -4(a - b)2

Since square quantity is always positive

So, (a - b)2 ≥ 0

So, D = -4(a - b)2 will be negative.

Hence the equation has no real roots.

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