sin alpha and cos alpha are the roots of the equation ax2 +bx +c then the value of b=?
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ax^2+bx+c = 0
x1 = [-b+(b^2–4ac)^0.5]/2a = sin alpha
x2 = [-b-(b^2–4ac)^0.5]/2a = cos alpha
But sin^2 alpha + cos^2 alpha = 1, or
{[-b+(b^2–4ac)^0.5]/2a}^2 + {[-b-(b^2–4ac)^0.5]/2a}^2 = 1, or
b^2 -2b(b^2–4ac)^0.5 + (b^2–4ac) + b^2 +2b(b^2–4ac)^0.5 + (b^2–4ac) = 4a^2
2b^2+2b^2–8ac = 4a^2
4b^2 = 4a^2+8ac
b^2 = a^2+2ac. Answer
x1 = [-b+(b^2–4ac)^0.5]/2a = sin alpha
x2 = [-b-(b^2–4ac)^0.5]/2a = cos alpha
But sin^2 alpha + cos^2 alpha = 1, or
{[-b+(b^2–4ac)^0.5]/2a}^2 + {[-b-(b^2–4ac)^0.5]/2a}^2 = 1, or
b^2 -2b(b^2–4ac)^0.5 + (b^2–4ac) + b^2 +2b(b^2–4ac)^0.5 + (b^2–4ac) = 4a^2
2b^2+2b^2–8ac = 4a^2
4b^2 = 4a^2+8ac
b^2 = a^2+2ac. Answer
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