How do you evaluate $\frac{8 + \sqrt{3^{2} + 2 \cdot 3 + 1}}{- 4}$ $(8+ \sqrt { 3^ { 2} + 2\cdot 3+ 1})/-4$ ?

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Answer 1 · 2018-08-04T06:53:45

$- 3$ $-3$

$\frac{8 + \sqrt{3^{2} + 2 \cdot 3 + 1}}{- 4}$ $(8 + sqrt(3^2 + 2 * 3 + 1))/-4$

$= \frac{8 + \sqrt{9 + 6 + 1}}{- 4}$ $=(8 + sqrt(9 + 6 + 1))/-4$

$= \frac{8 + \sqrt{16}}{- 4}$ $=(8 + sqrt(16))/-4$

$= \frac{8 + 4}{- 4}$ $=(8+4)/-4$

$= \frac{12}{- 4}$ $=12/-4$

$= - 3$ $=-3$

Hope this helps!

How do you evaluate 8+√32+2⋅3+1−4(8+ \sqrt { 3^ { 2} + 2\cdot 3+ 1})/-4?