# If sin α and cos α are the roots of the equation ax2 + bx + c = 0, then b2 =A. a2 – 2acB. a2 + 2acC. a2 – acD. a2 + ac

Equation ax2 + bx + c = 0 has and as two roots

sin α + cos α = sin α × cos α = c/a …eq(1)  …..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.

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