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# help Algebra

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x^2 + bx + b + 3 = 0 has roots of the form x = (-b +/- sqrt(7))/2, where b > 0.  Find b.

Nov 17, 2021

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$$b^2 - 4ac = 7$$

$$b^2 - 4(1)(b + 3) = 7$$
$$b^2 - 4b - 12 = 7$$
$$b^2 - 4b - 19 = 0$$
$$b = {-(-4) \pm \sqrt{(-4)^2-4(1)(-19)} \over 2(1)}$$

$$b = {4 \pm \sqrt{16+ 76} \over 2}$$

$$b = {4 \pm 2\sqrt{23} \over 2}$$

$$b = 2\pm\sqrt{23}$$

$$b>0$$

$$\therefore b = 2 + \sqrt{23}$$

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Nov 17, 2021