Question

SOMEONE HELP ME IM FREAKING OUT I LITERALLY CANT WITH THIS QUESTION IM PRAYING PLEASE HELP ME IM SO SERIOUS IM GONNA END IT PLS HELP ME the question is like 3-v BUT LIKE THE v has a line on top so its like 3-v32/1+v2 LIKE I DONT UNDERSTAND
can be written in the form a + bv2
where a and b are integers .

Answer 1

Answer:

$\sf -11+7\sqrt{2}$

Step-by-step explanation:

Given expression:

$\sf \dfrac{3-\sqrt{32}}{1+\sqrt{2} }$

Rewrite 32 as 16 · 2:

$\sf \implies \dfrac{3-\sqrt{16 \cdot 2}}{1+\sqrt{2} }$

Apply radical rule $\sf \sqrt{a \cdot b}=\sqrt{a}\sqrt{b}$

$\sf \implies \dfrac{3-\sqrt{16}\sqrt{2}}{1+\sqrt{2} }$

As $\sf \sqrt{16}=4$:

$\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} }$

Multiply by the conjugate:

$\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} } \times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }$

$\sf \implies \dfrac{(3-4\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}$

$\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4\sqrt{2}\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}$

As $\sf \sqrt{2}\sqrt{2}=\sqrt{4}=2$:

$\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4 \cdot 2}{1-\sqrt{2}+\sqrt{2}-2}$

$\sf \implies \dfrac{3-7\sqrt{2}+8}{1-2}$

$\sf \implies \dfrac{11-7\sqrt{2}}{-1}$

$\sf \implies -11+7\sqrt{2}$