x^2 + bx + b + 3 = 0 has roots of the form x = (-b +/- sqrt(7))/2, where b > 0. Find b.
+592 \(b^2 - 4ac = 7\)
\(b^2 - 4(1)(b + 3) = 7\)
\(b^2 - 4b - 12 = 7\)
\(b^2 - 4b - 19 = 0\)
\(b = {-(-4) \pm \sqrt{(-4)^2-4(1)(-19)} \over 2(1)}\)
\(b = {4 \pm \sqrt{16+ 76} \over 2}\)
\(b = {4 \pm 2\sqrt{23} \over 2}\)
\(b = 2\pm\sqrt{23}\)
\(b>0 \)
\(\therefore b = 2 + \sqrt{23}\)
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