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qaws [65]
4 years ago
10

What are the solutions of the equation x6 + 6x3 + 5 = 0? Use factoring to solve.

Mathematics
1 answer:
lapo4ka [179]4 years ago
6 0

Consider the equation x^6+6x^3+5=0.

First, you can use the substitution t=x^3, then x^6=(x^3)^2=t^2 and equation becomes t^2+6t+5=0. This equation is quadratic, so

D=6^2-4\cdot 5\cdot 1=36-20=16=4^2,\ \sqrt{D}=4,\\ \\ t_{1,2}=\dfrac{-6\pm 4}{2} =-5,-1.

Then you can factor this equation:

(t+5)(t+1)=0.

Use the made substitution again:

(x^3+5)(x^3+1)=0.

You have in each brackets the expression like a^3+b^3 that is equal to (a+b)(a^2-ab+b^2). Thus,

x^3+5=(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25}) ,\\x^3+1=(x+1)(x^2-x+1)

and the equation is

(x+\sqrt[3]{5})(x^2-\sqrt[3]{5}x+\sqrt[3]{25})(x+1)(x^2-x+1)=0.

Here x_1=-\sqrt[3]{5} , x_2=-1 and you can sheck whether quadratic trinomials have real roots:

1. D_1=(-\sqrt[3]{5}) ^2-4\cdot \sqrt[3]{25}=\sqrt[3]{25} -4\sqrt[3]{25} =-3\sqrt[3]{25}.

2. D_2=(-1)^2-4\cdot 1=1-4=-3.

This means that quadratic trinomials don't have real roots.

Answer: x_1=-\sqrt[3]{5} , x_2=-1

If you need complex roots, then

x_{3,4}=\dfrac{\sqrt[3]{5}\pm i\sqrt{3\sqrt[3]{25}}}{2}   ,\\ \\x_{5,6}=\dfrac{1\pm i\sqrt{3}}{2}.

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3 years ago
(-6,8); perpendicular to y = -3/2x -1
UNO [17]

Answer:

y =  \frac{2}{3} x + 12

Step-by-step explanation:

y =  -  \frac{3}{2} x - 1

The gradient of a line is the coefficient of x when the equation of the line is written in the form of y=mx+c.

Thus, gradient of given line=-  \frac{3}{2}.

The product of the gradients of perpendicular lines is -1.

(Gradient of line)(-3/2)= -1

Gradient of line

- 1 \div ( -  \frac{3}{2} ) \\  =  - 1( -  \frac{2}{3} )  \\  =  \frac{2}{3}

Substitute m=\frac{2}{3} into y=mx+c:

y =  \frac{2}{3} x + c

To find the value of c, substitute a pair of coordinates.

When x= -6, y= 8,

8 =  \frac{2}{3} ( - 6) + c \\  \\ 8 =  - 4 + c \\ c = 8 + 4 \\ c = 12

Thus, the equation of the line is y =  \frac{2}{3} x + 12.

7 0
3 years ago
Paula has a dog that weighs 3 times as much as Carla's dog
ivanzaharov [21]

Answer:

The weight of Paula's dog is 36 pounds.

Step-by-step explanation:

SO we would draw one box for the weight of Carla's dog. Then we would draw 3 boxes for the weight of Paula's dog.

To find how much does Paula's dog weigh we divide the total weight of the dogs by the total number of boxes.

So then you would calculate 48 divided by 4.

The weight of Carla's dog is 12 pounds.

Therefore the weight of Paula's dog is 3x12=3x10+3x2=30+6=36 pounds.

4 0
3 years ago
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