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3 years ago

Simplify fully (square root symbol)12

Mathematics

1 answer:

denpristay [2]3 years ago

4 0

Look at the attached picture⤴

Hope it will help u...

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Solve by completing the square x^2-4x+9=0

Masteriza [31]

Step-by-step explanation:

Add 9 to both sides of the equation.

if you want to create a trinomial square on the left, find value that is equal to the square of half of b

Add the term to each side of the equation

SIMPLIFY >:(

How to Simplify it:

-Raise the -2 to the power of 2

and ten simplify 9 + (-2)^2 :

--Raise -2 to the power of 2 and then add 9 and 4

Factor the perfect trinomial square into (x - 2)^2

Solve the equation for x

How to do it:

-Take the square root of each side of the equation to set up the solution for x

--Then remove the perfect root factor, x - 2 under the radical to solve for x

---Remove parenthesis

----add 2 to both sides of the equation.

the final answer could be shown in different ways,but the exact form is x=±√13+2

In decimal form its x=5.60555127…,−1.60555127…

--this took me like 7 minutes to do im no geek tho lol

7 0

3 years ago

Find the cube roots of 27(cos 330° + i sin 330°)

Aleksandr-060686 [28]

Answer:

See below for all the cube roots

Step-by-step explanation:

DeMoivre's Theorem

Let $z=r(cos\theta+isin\theta)$ be a complex number in polar form, where $n$ is an integer and $n\geq1$ . If $z^n=r^n(cos\theta+isin\theta)^n$ , then $z^n=r^n(cos(n\theta)+isin(n\theta))$ .

Nth Root of a Complex Number

If $n$ is any positive integer, the nth roots of $z=rcis\theta$ are given by $\sqrt[n]{rcis\theta}=(rcis\theta)^{\frac{1}{n}}$ where the nth roots are found with the formulas:

$\sqrt[n]{r}\biggr[cis(\frac{\theta+360^\circ k}{n})\biggr]$ for degrees (the one applicable to this problem)
$\sqrt[n]{r}\biggr[cis(\frac{\theta+2\pi k}{n})\biggr]$ for radians

for $k=0,1,2,...\:,n-1$

Calculation

$z=27(cos330^\circ+isin330^\circ)\\\\\sqrt[3]{z} =\sqrt[3]{27(cos330^\circ+isin330^\circ)}\\\\z^{\frac{1}{3}} =(27(cos330^\circ+isin330^\circ))^{\frac{1}{3}}\\\\z^{\frac{1}{3}} =27^{\frac{1}{3}}(cos(\frac{1}{3}\cdot330^\circ)+isin(\frac{1}{3}\cdot330^\circ))\\\\z^{\frac{1}{3}} =3(cos110^\circ+isin110^\circ)$

First cube root where k=2

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(2)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+720^\circ}{3})\biggr]\\3\biggr[cis(\frac{1050^\circ}{3})\biggr]\\3\biggr[cis(350^\circ)\biggr]\\3\biggr[cos(350^\circ)+isin(350^\circ)\biggr]$

Second cube root where k=1

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(1)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ+360^\circ}{3})\biggr]\\3\biggr[cis(\frac{690^\circ}{3})\biggr]\\3\biggr[cis(230^\circ)\biggr]\\3\biggr[cos(230^\circ)+isin(230^\circ)\biggr]$

Third cube root where k=0

$\sqrt[3]{27}\biggr[cis(\frac{330^\circ+360^\circ(0)}{3})\biggr]\\3\biggr[cis(\frac{330^\circ}{3})\biggr]\\3\biggr[cis(110^\circ)\biggr]\\3\biggr[cos(110^\circ)+isin(110^\circ)\biggr]$