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thank you!

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If $x^2 + bx + b + 3 = 0$ has roots of the form $\frac{-b \pm \sqrt{5}}{2}$, where $b > 0$, then $b = m+\sqrt{n}$ for positive integers $m,n$. Find $m + n$.

May 6, 2018

#1
+22581
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If $x^2 + bx + b + 3 = 0$ has roots of the form $\frac{-b \pm \sqrt{5}}{2}$, where $b > 0$,

then $b = m+\sqrt{n}$ for positive integers $m,n$. Find $m + n$.

$$\begin{array}{|rcll|} \hline \left( x - \frac{-b +\sqrt{5}}{2} \right)\left( x - \frac{-b - \sqrt{5}}{2} \right) &=& x^2+bx+b+3 \\ \left( x + \frac{b}{2} -\frac{\sqrt{5}}{2} \right)\left( x + \frac{b}{2} +\frac{\sqrt{5}}{2} \right) &=& x^2+bx+b+3 \\ \left( ( x + \frac{b}{2}) -\frac{\sqrt{5}}{2} \right)\left( (x + \frac{b}{2}) +\frac{\sqrt{5}}{2} \right) &=& x^2+bx+b+3 \\ \left( x + \frac{b}{2} \right)^2- \left(\frac{\sqrt{5}}{2} \right)^2 &=& x^2+bx+b+3 \\ x^2+xb+\frac{b^2}{4}- \frac{5}{4} &=& x^2+bx+b+3 \\ \frac{b^2}{4}- \frac{5}{4} &=& b+3 \\ \frac{b^2}{4}-b - \frac{5}{4} -3 &=& 0 \quad & | \quad \cdot 4 \\ b^2-4b - 5 -12 &=& 0 \\ b^2-4b - 17 &=& 0 \\\\ b &=& \frac{4\pm \sqrt{16-4\cdot(-17)}}{2} \\ &=& \frac{4\pm \sqrt{16+68}}{2} \\ &=& \frac{4\pm \sqrt{84}}{2} \\ &=&2\pm \frac{\sqrt{84}}{2} \\ &=&2\pm \sqrt{\frac{84}{4}} \\ &=&2\pm \sqrt{21} \quad & | \quad b>0\ !\\\\ \mathbf{b} & \mathbf{=} & \mathbf{2+\sqrt{21} } \\ b &=& m+\sqrt{n} \quad & | \quad m = 2 \text{ and } n = 21 \\ \mathbf{m+n} &\mathbf{=}& \mathbf{2+21=23} \\ \hline \end{array}$$

May 7, 2018
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