Show that the equation \(2(a^2+b^2)\text{x}^2+2(a+b)\text{x}+1=0\) h

1.	Show that the equation \(2(a^2+b^2)\text{x}^2+2(a+b)\text{x}+1=0\) has no real roots, when a ≠ b.
Answer» For a quadratic equation, ax²+ bx + c = 0, D = b² – 4ac If D < 0, roots are not real. \(2(a^2+b^2)\text{x}^2+2(a+b)\text{x}+1=0\) ⇒ D = 4(a + b)² – 8(a² + b²) ⇒ D = 4a² + 4b² + 8ab – 8a² – 8b² ⇒ D = - 4(a² + b² – 2ab) = - 4(a – b)² Thus, D < 0 for all values of a and b. ∴ Roots are not real.

Show that the equation \(2(a^2+b^2)\text{x}^2+2(a+b)\text{x}+1=0\) has no real roots, when a ≠ b.

Answer»

For a quadratic equation, ax²+ bx + c = 0,

D = b² – 4ac

If D < 0, roots are not real.

\(2(a^2+b^2)\text{x}^2+2(a+b)\text{x}+1=0\)

⇒ D = 4(a + b)² – 8(a² + b²)

⇒ D = 4a² + 4b² + 8ab – 8a² – 8b²

⇒ D = - 4(a² + b² – 2ab) = - 4(a – b)²

Thus, D < 0 for all values of a and b.

∴ Roots are not real.