12th q
Q.12. Show that the equation 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.