Question

What is (root 3 + root 2 - 1) multiplied by (root 3 + root 2) + 1

Answer 1

☯ To FinD :

We have to find the value of (√3 + √2 -1 )* (√3 + √2 + 1)

$\rule{200}{2}$

☯ SolutioN :

If we see the question clearly then we keen to be know that we can use identity in this question.

So,

Let (√3 + √2) be a and (1) be b.

Now, we know that,

$\Large{\implies{\boxed{\boxed{\sf{(a + b)(a - b) = a^2 - b^2}}}}}$

$\sf{\dashrightarrow (\sqrt{3} + \sqrt{2})^2 - (1)^2} \\ \\ \bf{Now,} \\ \\ \Large{\star{\boxed{\sf{(a + b)^2 = (a)^2 + (b)^2 + 2ab}}}} \\ \\ \sf{\dashrightarrow (\sqrt{3})^2 + (\sqrt{2})^2 + 2(\sqrt{3})(\sqrt{2}) - 1} \\ \\ \sf{\dashrightarrow 3 + 2 - 1 + \sqrt{6}} \\ \\ \sf{\dashrightarrow 4 + 2\sqrt{6}} \\ \\ \Large{\implies{\boxed{\boxed{\sf{4 + 2 \sqrt{6} }}}}}$

Answer 2

$(\begin{gathered}\sf{\dashrightarrow (\sqrt{3} + \sqrt{2})^2 - (1)^2} \\ \\ \bf{Now,} \\ \\ \Large{\star{\boxed{\sf{(a + b)^2 = (a)^2 + (b)^2 + 2ab}}}} \\ \\ \sf{\dashrightarrow (\sqrt{3})^2 + (\sqrt{2})^2 + 2(\sqrt{3})(\sqrt{2}) - 1} \\ \\ \sf{\dashrightarrow 3 + 2 - 1 + \sqrt{6}} \\ \\ \sf{\dashrightarrow 4 + 2\sqrt{6}} \\ \\ \Large{\implies{\boxed{\boxed{\sf{4 + 2 \sqrt{6} }}}}}\end{gathered}</p><p>$