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19.10.2020 •

Mathematics

Describe the solutions to the quadratic equation below: x^2+6x+9=0 2 irrational roots
2 complex roots
1 rational (double) root
2 rational roots

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Ответ:

chandranewlon

30.05.2020

Mathematics

$BE = 4\sqrt2$

Step-by-step explanation:

Given

The attached square

Required

Find BE

First, we calculate BD (the diagonal) using Pythagoras theorem.

BD^2 = BA^2 + AD^2

Since the shape is a square, then

BA = AD = 8

So:

BD^2 = 8^2 + 8^2

BD^2 = 2(8^2)

Take square roots

$BD = \sqrt{2(8^2)$

$BD = \sqrt{2} * 8$

$BD = 8\sqrt{2}$

BE is half BD.

So:

$BE = \frac{1}{2} * BD$

$BE = \frac{1}{2} * 8\sqrt2$

$BE = 4\sqrt2$

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Describe the solutions to the quadratic equation below: x^2+6x+9=0 2 irrational roots 2 complex roots 1 rational (double) root 2 rational roots