How do you simplify (sqrt5-3)^2?

Nov 20, 2017

$14 - 6 \sqrt{5}$

Explanation:

Recall the identity for a binomial squared:

${\left(a + b\right)}^{2} = {a}^{2} + {b}^{2} + 2 a b$

So, all your doing is using this formula, but with two numbers. This gives you:

${\left(\sqrt{5} - 3\right)}^{2} = {\left(\sqrt{5}\right)}^{2} + {\left(- 3\right)}^{2} + 2 \left(\sqrt{5} \left(- 3\right)\right)$

$= 5 + 9 - 6 \sqrt{5}$

$\implies 14 - 6 \sqrt{5}$

However, there is also another formula you might stumble across that you could use:

${\left(a - b\right)}^{2} = {a}^{2} + {b}^{2} - 2 a b$

Can you use this and get the same answer? YES! This is exactly the same thing . Instead of plugging a negative number into the first formula, we just take care of that by making $b$ negative beforehand, and then dealing purely in positive numbers.

Either way, you should end up at the same answer.

Hope that helped :)

Nov 20, 2017

$= 2 \left(7 - 3 \sqrt{5}\right)$

Explanation:

Use identity: ${\left(x - y\right)}^{2} = {x}^{2} - 2 x y + {y}^{2}$

${\left(\sqrt{5} - 3\right)}^{2}$----- here $x = \sqrt{5} \mathmr{and} y = 3$

$\implies {\left(\sqrt{5} - 3\right)}^{2} = {\left(\sqrt{5}\right)}^{2} - 2 \times \sqrt{5} \times 3 + {\left(3\right)}^{2}$

$\implies {\left(\sqrt{5} - 3\right)}^{2} = 5 - 6 \sqrt{5} + 9$

$\implies {\left(\sqrt{5} - 3\right)}^{2} = 14 - 6 \sqrt{5}$

${\left(\sqrt{5} - 3\right)}^{2} = 2 \left(7 - 3 \sqrt{5}\right)$

Nov 20, 2017

$5 - 6 \sqrt{5} + 9$

Explanation:

First, let $\sqrt{5}$ be $a$ and $- 3$ be b.

Our simplified answer will be in the form of ${a}^{2} + 2 a b + {b}^{2}$.

First, to get ${a}^{2}$, we do this: ${\sqrt{5}}^{2}$ means that the sqrt and squared get cancelled. So that becomes $5$.

Then we need $2 a b$. $2 \left(- 3\right) \left(\sqrt{5}\right) = - 6 \sqrt{5}$

Finally, we need ${b}^{2}$, which is ${\left(- 3\right)}^{2} \to 9$

So let's put all of these together, so our final answer is:
$5 - 6 \sqrt{5} + 9$