What is the square root of 405? and explain it

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What is the square root of 405? and explain it

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Answer 1 · 2018-03-07T18:23:50

Find the two perfect square roots closest to 405:

$20^{2} = 400$ $20^2=400$

$21^{2} = 441$ $21^2=441$

Write an equation using this info, with the points as $(perfect square, square root of that perfect square)$ $("perfect square", "square root of that perfect square")$ :

$(400, 20), (441, 21)$ $(400,20), (441,21)$

Make an equation by finding the slope and the y-int:

$\frac{21 - 20}{441 - 400} = \frac{1}{41}$ $(21-20)/(441-400)=1/41$

$y = \frac{1}{41} x + b$ $y=1/41x+b$

$20 = \frac{1}{41} \cdot 400 + b$ $20=1/41*400+b$

$b = 10.24390$ $b=10.24390$

$y = 0.024390 x + 10.24390$ $y=0.024390x+10.24390$

Plug in $405$ $405$ as $x$ $x$ :

$y = 0.024390 \cdot 405 + 10.24390 \approx 20.09$ $y=0.024390*405+10.24390~~20.09$

About $20.09$ $20.09$

Approximately only, not exact.

Answer 2 · 2018-03-07T19:11:22

$\sqrt{405} = 9 \sqrt{5} \approx 20.1246118$ $sqrt(405)=9sqrt(5)~~20.1246118$

Explanation:

Extracting the obvious factor of $5$ $5$ from $405$ $405$ , we get
$405 = 5 \cdot 81$ $405=5 * 81$
Assuming we recognize $81$ $81$ as a perfect square ( $= 9^{2}$ $=9^2$ ), we can then write
$405 = 5 \cdot 9^{2}$ $405 = 5 * 9^2$

then
$\sqrt{405} = \sqrt{5 \cdot 9^{2}}$ $sqrt(405)=sqrt(5 * 9^2)$
$XXX = \sqrt{5} \cdot \sqrt{9^{2}}$ $color(white)("XXX")=sqrt(5) * sqrt(9^2)$
$XXX = \sqrt{5} \cdot 9$ $color(white)("XXX")=sqrt(5) * 9$
$XXX = 9 \sqrt{5}$ $color(white)("XXX")=9sqrt(5)$

If you need to find an approximation without the radical, you would likely want to use a calculator (there is a "paper-and-pencil" method but it is seldom used/taught).

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What is the square root of 405? and explain it

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