Equation ax2 + bx + c = 0 has cos a and cos a as two roots
sin α + cos α = \(-\frac{b}{a}\)
sin α × cos α = c/a …eq(1)
\((sin\,a\,+\,cosa)^2=\frac{b^2}{a^2}\)
\(sin^2a\,+cos^2a\,+2\,sina\,cos=\frac{b^2}{a^2}....eq(2)\)
But sin2 α + cos2 α = 1
∴ a2 (1 + 2 sinα.cos α) = b2
Putting sin α × cos α = c/a, we get,
⇒ b2 = a2 + 2ac.