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3 years ago

Solve the equation in the complex number system. x^2-12x+40=0

Mathematics

1 answer:

mylen [45]3 years ago

5 0

solving the equation $x^2-12x+40=0$ we get $x=6+2i\,\,and\,\,x=6-2i$

Step-by-step explanation:

Solve the equation:

$x^2-12x+40=0$

We can solve using quadratic Formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b = -12, c = 40

Putting values:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(40)}}{2(1)}\\x=\frac{12\pm\sqrt{144-160}}{2}\\x=\frac{12\pm\sqrt{-16}}{2}\\We\,\,know\,\,\sqrt{-1}=i\\ x=\frac{12\pm4i}{2}\\ x=\frac{12+4i}{2}\,\,and\,\, x=\frac{12-4i}{2}\\ x=\frac{4(3+i)}{2}\,\,and\,\, x=\frac{4(3-i)}{2}\\x=2(3+i)\,\,and\,\,x=2(3-i)\\x=6+2i\,\,and\,\,x=6-2i$

So, solving the equation $x^2-12x+40=0$ we get $x=6+2i\,\,and\,\,x=6-2i$

Keywords: Solving quadratic equation

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Evaluate the definite integral. <br> 1 x4(1 + 2x5)5 dx.

Nata [24]

Answer:

Step-by-step explanation:

Given the definite integral $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ , we to evaluate it. Using integration by substitution method.

Let u = 1-2x⁵ ...1

du/dx = -10x⁴

dx = du/-10x⁴.... 2

Substitute equation 1 and 2 into the integral function and evaluate the resulting integral as shown;

$= \int\limits {\dfrac{x^4}{u^5} } \, \dfrac{du}{-10x^4}$

$= \dfrac{-1}{10} \int\limits {\dfrac{du}{u^5} } \\\\= \dfrac{-1}{10} \int\limits {{u^{-5}du } \\= \dfrac{-1}{10} [{\frac{u^{-5+1}}{-5+1}] \\\\= \dfrac{-1}{10} ({\frac{u^{-4}}{-4})\\\\$

$= \dfrac{u^{-4}}{40} \\\\\\= \dfrac{1}{40u^4} +C$

substitute u = 1-2x⁵ into the result

$= \dfrac{1}{40(1-2x^5)^4} +C$

Hence $\int\limits {\dfrac{x^4}{(1-2x^5)^5} } \, dx$ $= \dfrac{1}{40(1-2x^5)^4} +C$

5 0

3 years ago

Solve x2 - 8x - 9 = 0.<br><br> Rewrite the equation so that it is of the form <br> x2 + bx = c.

stira [4]

X^2 - 8x-9=0
D = b^2- 4*a*c = 64- 4* (-9) = 64+36=100 =10^2
X1 = (8-10)/2 = -1
X2= (8+10)/2= 9

7 0

3 years ago

Reflect A (1,3) over the y-axis

gulaghasi [49]

Answer:

Up 1 and over 3

Step-by-step explanation:

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3 years ago

HELP!!!

Bad White [126]

I hope this helps you

-15+6= -9

-9+6= -3

-3+6=3

add 6

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3 years ago

Read 2 more answers

Square root of 32 x square root of 1 over 18

lisov135 [29]

<h3>Solution:</h3>

$\sqrt{32} \times \sqrt{ \frac{1}{18} } \\ = \sqrt{32} \times \frac{ \sqrt{1} }{ \sqrt{18} } \\ = \sqrt{32} \times \frac{1}{ \sqrt{18} } \\ = \frac{\sqrt{2 \times 2 \times 2 \times 2 \times 2}}{ \sqrt{2 \times 3 \times 3} } \\ = \frac{ \sqrt{ {2}^{2} \times {2}^{2} \times 2} }{ \sqrt{2 \times {3}^{2} } } \\ = \frac{2 \times 2 \times \sqrt{2} }{3 \times \sqrt{2} } \\ = \frac{4 \times \sqrt{2} }{3 \times \sqrt{2} } \\ = \frac{4}{3}$

<h3>Answer:</h3>

$\frac{4}{3}$

<h3>Hope it helps...</h3>

ray4918 here to help

7 0

2 years ago