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3 years ago

What are the solutions of x^2-2x+10=0

2 answers:

8 0

There are a few ways to solve this, the simplest of which is probably factoring. Start by finding factors of C (10) that add up to be B (-2).

fac(10) { -10, 1 ; -5, 2 }

Well, there are no factors of 10 that can add to b - so we will use the quadratic formula.

(-b +- sqrt(b^2 - 4ac)) / 2a

where a = 1, b = -2, c = 10

b^2 - 4ac = (-2)^2 - 4(1)(10) = 4 - 40 = -36

-b +- sqrt(-36) / 2a

(2 +- sqrt(-36)) / 2

sqrt(-36)

sqrt(-1 * 36)

sqrt(-1) * sqrt(36)

i * 6

2 + 6i / 2 ; 1 + 3i

2 - 6i / 2 ; 1 - 3i

Solutions: 1 + 3i, 1 - 3i

cluponka [151]3 years ago

7 0

Answer:

x = 1± 3i

Step-by-step explanation:

x^2-2x+10=0

We can complete the square to solve by subtracting 10 from each side

x^2-2x+10-10=-10

x^2 -2x = -10

We need to add (2/2) ^2 to each side or 1

x^2 -2x+1 = -10 +1

x^2 -2x+1 = -9

The left side factors into (x- (2/2) ) ^2

(x-1) ^2 = -9

Take the square root of each side

sqrt((x-1) ^2 =± sqrt(-9)

x-1 = ±sqrt(-1) sqrt(3)

Remember the sqrt(-1) = i

x-1 = ± 3i

Add 1 to each side

x-1+1 = 1± 3i

x = 1± 3i

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Find all the complex roots. Write the answer in exponential form.

dezoksy [38]

We have to calculate the fourth roots of this complex number:

$z=9+9\sqrt[]{3}i$

We start by writing this number in exponential form:

$\begin{gathered} r=\sqrt[]{9^2+(9\sqrt[]{3})^2} \\ r=\sqrt[]{81+81\cdot3} \\ r=\sqrt[]{81+243} \\ r=\sqrt[]{324} \\ r=18 \end{gathered}$ $\theta=\arctan (\frac{9\sqrt[]{3}}{9})=\arctan (\sqrt[]{3})=\frac{\pi}{3}$

Then, the exponential form is:

$z=18e^{\frac{\pi}{3}i}$

The formula for the roots of a complex number can be written (in polar form) as:

$z^{\frac{1}{n}}=r^{\frac{1}{n}}\cdot\lbrack\cos (\frac{\theta+2\pi k}{n})+i\cdot\sin (\frac{\theta+2\pi k}{n})\rbrack\text{ for }k=0,1,\ldots,n-1$

Then, for a fourth root, we will have n = 4 and k = 0, 1, 2 and 3.

To simplify the calculations, we start by calculating the fourth root of r:

$r^{\frac{1}{4}}=18^{\frac{1}{4}}=\sqrt[4]{18}$

NOTE: It can not be simplified anymore, so we will leave it like this.

Then, we calculate the arguments of the trigonometric functions:

$\frac{\theta+2\pi k}{n}=\frac{\frac{\pi}{2}+2\pi k}{4}=\frac{\pi}{8}+\frac{\pi}{2}k=\pi(\frac{1}{8}+\frac{k}{2})$

We can now calculate for each value of k:

$\begin{gathered} k=0\colon \\ z_0=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{0}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{0}{2}))) \\ z_0=\sqrt[4]{18}\cdot(\cos (\frac{\pi}{8})+i\cdot\sin (\frac{\pi}{8}) \\ z_0=\sqrt[4]{18}\cdot e^{i\frac{\pi}{8}} \end{gathered}$ $\begin{gathered} k=1\colon \\ z_1=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{1}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{1}{2}))) \\ z_1=\sqrt[4]{18}\cdot(\cos (\frac{5\pi}{8})+i\cdot\sin (\frac{5\pi}{8})) \\ z_1=\sqrt[4]{18}e^{i\frac{5\pi}{8}} \end{gathered}$ $\begin{gathered} k=2\colon \\ z_2=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{2}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{2}{2}))) \\ z_2=\sqrt[4]{18}\cdot(\cos (\frac{9\pi}{8})+i\cdot\sin (\frac{9\pi}{8})) \\ z_2=\sqrt[4]{18}e^{i\frac{9\pi}{8}} \end{gathered}$ $\begin{gathered} k=3\colon \\ z_3=\sqrt[4]{18}\cdot(\cos (\pi(\frac{1}{8}+\frac{3}{2}))+i\cdot\sin (\pi(\frac{1}{8}+\frac{3}{2}))) \\ z_3=\sqrt[4]{18}\cdot(\cos (\frac{13\pi}{8})+i\cdot\sin (\frac{13\pi}{8})) \\ z_3=\sqrt[4]{18}e^{i\frac{13\pi}{8}} \end{gathered}$

Answer:

The four roots in exponential form are

z0 = 18^(1/4)*e^(i*π/8)

z1 = 18^(1/4)*e^(i*5π/8)

z2 = 18^(1/4)*e^(i*9π/8)

z3 = 18^(1/4)*e^(i*13π/8)

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Maksim231197 [3]

Answer:

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