Correct option is (B) \(b^2-2ac=a^2\)
Given quadratic equation is
\(ax^2+bx+c=0\)
Given that \(sin\;\alpha\) and \(cos\;\alpha\) are roots of the equation
\(ax^2+bx+c=0\)
\(\therefore\) Sum of roots \(=\frac{-b}a\)
\(\Rightarrow\) \(sin\;\alpha+cos\;\alpha\) \(=\frac{-b}a\) ______________(1)
& Product of roots \(=\frac ca\)
\(\Rightarrow\) \(sin\;\alpha\;cos\;\alpha\) \(=\frac ca\) ______________(2)
\(\because\) \(sin^2\alpha+cos^2\alpha=1\)
\(\Rightarrow\) \((sin\;\alpha+cos\;\alpha)^2\) \(-2\;sin\;\alpha\;cos\;\alpha=1\)
\(\Rightarrow(\frac{-b}a)^2-\frac{2c}a=1\)
\(\Rightarrow\) \(\frac{b^2-2ac}{a^2}=1\)
\(\Rightarrow\) \(b^2-2ac=a^2\)
