What is 40 times the square root of 6?

Sep 18, 2015

$40 \sqrt{6}$ is the most reduced form you get. You can't go any further.

Now if you were to approximate $\sqrt{6}$, you can approximate an answer for $40 \sqrt{6}$.

$\sqrt{4} = 2$
sqrt6 = ?
$\sqrt{9} = 3$

If you assume that some % between ${x}_{1}^{2}$ and ${x}_{2}^{2}$ is the same as the % between ${x}_{1}$ and ${x}_{2}$, you can say that:

$\frac{6 - 4}{9 - 4} \cdot \left(3 - 2\right) + 2 = 2.4$ (or 40% of the way between $2$ and $3$)

is the approximate value of $\sqrt{6}$, which would give $40 \sqrt{6} \approx 40 \cdot 2.4 = 96$. It's an underestimate though, since what we are really doing is drawing a straight line with a slope of $5$ coming from $y = 4$ and intersecting at $x = 2$ and $x = 3$, and this line is to the left of the actual ${x}^{2}$ curve at $y = 6$:

graph{(y-x^2)(y + 6 - 5x) = 0 [-1.5, 4, -0.5, 10]}

In actuality it is about $2.45$, and $40 \sqrt{6} \approx 97.98$.