Question

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

prove that

a² - b² + 2ac = 0

OR

a² + 2ac = b² ​

Answer 1

$\huge\orange{\boxed{\bold{Solution}}}$

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$\bold {ax}^{2} + \bold {bx} \: + \bold c \: = \bold 0 \\ \\ \bold {sin (\alpha) \times cos (\alpha)} = \bold {\frac{ - b}{a} } \\ \\ \bold {sin (\alpha) \times cos (\alpha)} = \bold {\frac{c}{a}} \\ \\ \bold {[sin (\alpha) + cos (\alpha) ]^{2}} = \bold {{sin}^{2} (\alpha) + 2sin \times cos (\alpha) + \: {cos}^{2} (\alpha)} \\ = \bold {{sin}^{2} (\alpha) + {cos}^{2} (\alpha) + 2sin \times cos (\alpha)} \\ \\ \bold {( \frac{ - b}{a})^{2}} = \bold {1 + \frac{2c}{a}} \\ \\ \bold {\frac{ {b}^{2} }{ {a}^{2} }} \: = \bold {1 + \frac{2c}{a}} \\ \\ \bold {{b}^{2}} = \bold {\frac{ {a}^{2}(a + 2c) }{a}} \\ \\ \bold {{b}^{2}} = \: \bold {{a}^{2}} + \: \bold {2ac}$

$\huge\underline\mathfrak\green{Hence\:proved}$

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$\huge\mathbb\blue{THANK\:YUH!}$

Answer 2

see solution in image ..

Hope it helps you .