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(B) We have,$x^2+|x-3|=4$. Case $I$: $x \ge 3$. The equation becomes $x^2 + x - 3 = 4$,which simplifies to $x^2 + x - 7 = 0$. The roots are $x = \frac

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(B) We have,$x^2+|x-3|=4$. Case $I$: $x \ge 3$. The equation becomes $x^2 + x - 3 = 4$,which simplifies to $x^2 + x - 7 = 0$. The roots are $x = \frac{-1 \pm \sqrt{1 - 4(1)(-7)}}{2} = \frac{-1 \pm \sqrt{29}}{2}$. Since $\sqrt{29} \approx 5.38$,the roots are $\approx 2.19$ and $\approx -3.19$. Neither of these satisfies the condition $x \ge 3$. Case $II$: $x The equation becomes $x^2 - (x - 3) = 4$,which simplifies to $x^2 - x + 3 = 4$,or $x^2 - x - 1 = 0$. The roots are $x = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Both roots $\frac{1 + \sqrt{5}}{2} \approx 1.618$ and $\frac{1 - \sqrt{5}}{2} \approx -0.618$ satisfy the condition $x The sum of these roots is $\frac{1 + \sqrt{5}}{2} + \frac{1 - \sqrt{5}}{2} = \frac{2}{2} = 1$.

The sum of all the real numbers satisfying the equation $x^2+|x-3|=4$ is

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