If sin α and cos α are the roots of the equation ax2 +bx+c=0, then

. Since, sin α and cos α are the roots of the equation ax2+ bx+ c=0, then,sin⁡α+cos⁡α=−ba and sin⁡αcos⁡α=caTo eliminate α we have1=sin2⁡α+cos2⁡α⇒ 1=(sin⁡α+cos⁡α)2−2sin⁡αcos⁡α⇒ 1=b2a2−2ca⇒ a2−b2+2ac=0

If sin α and cos α are the roots of the equation ax2 +bx+c=0, then

. Since, sin α and cos α are the roots of the equation ax2+ bx+ c=0, then,sin⁡α+cos⁡α=−ba and sin⁡αcos⁡α=caTo eliminate α we have1=sin2⁡α+cos2⁡α⇒ 1=(sin⁡α+cos⁡α)2−2sin⁡αcos⁡α⇒ 1=b2a2−2ca⇒ a2−b2+2ac=0

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If sin $α$ and cos $α$ are the roots of the equation ax² +bx+c=0, then

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$a^{2} + b^{2} + 2 ac = 0$

$a^{2} + b^{2} - 2 ac = 0$

$(a - c)^{2} = b^{2} + c^{2}$

$a^{2} - b^{2} + 2 ac = 0$

answer is A.

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Detailed Solution

. Since, sin $α$ and cos $α$ are the roots of the equation ax²+ bx+ c=0, then,
$\sin α + \cos α = - \frac{b}{a} and \sin αcos α = \frac{c}{a}$
To eliminate $α$ we have
$1 = \sin^{2} α + \cos^{2} α$
$\begin{array}{ll} \Rightarrow 1 = (\sin α + \cos α)^{2} - 2 \sin αcos α \\ \Rightarrow 1 = \frac{b^{2}}{a^{2}} - \frac{2 c}{a} \\ \Rightarrow a^{2} - b^{2} + 2 ac = 0 \end{array}$