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2 years ago

What are the solutions of the equation x^4 – 5x^2 – 36 = 0? Use factoring to solve.

Mathematics

1 answer:

hammer [34]2 years ago

3 0

Answer:

Step-by-step explanation:

The equation is x^4 – 5x^2 – 36 = 0

We will break the middle term:

Firstly multiply the coefficient of x^4 by constant term of the equation:

1*36 = 36

Now find any two numbers whose product is 36 and their sum or difference is equal to 5

9*4 = 36

9-4 = 5

Now,

x^4 – 5x^2 – 36 = 0

x^4-9x^2+4x^2-36=0

Now take the common:

x^2(x^2-9)+4(x^2-9)=0

(x^2+4)(x^2-9)=0

x²+4=0,

x²= 0-4,

x²=-4,

Take root on both sides:

√x²=+/-√-4

+/-√-4 = +/-√-1 *√4

√-1 = i

Then +/-√-1 *√4 = √4 i

We know that the root of 4 is 2

Then we can write it as +/-2i

Thus x = 2i , -2i

Now (x^2-9)= 0

x²=0+9

x²=9

Take square root on both sides:

√x²=√9

x=+/-3

x= 3, -3

Therefore the values of x are 2i, -2i, 3 , -3 ....

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GenaCL600 [577]

Answer:

$(A)\ a=-6;b=4\\\\ (B)\ a=6;b=-4$

Step-by-step explanation:

The equation of the line in Slope-Intercept form is:

$y=mx+c$

Where "m" is the slope and "c" is the y-intercept.

By definition:

1. If the lines of the System of equations are parallel (whrn they have the same slope), the system has No solutions.

2. If the they are the same exact line, the System of equations has Infinite solutions.

(A) Let's solve for "y" from the first equation:

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

You can notice that:

$m=\frac{3}{2}\\\\c=\frac{1}{2}$

In order make that the System has No solutions, the slopes must be the same, but the y-intercept must not. Then, the values of "a" and "b" can be:

$a=-6\\\\b=4$

Substituting those values into the second equation and solving for "y", you get:

$-6x+4y=-2\\\\4y=6x-2\\\\y=\frac{6}{4}x-{2}{4}\\\\y=\frac{3}{2}x-\frac{1}{2}$

You can idenfity that:

$m=\frac{3}{2}\\\\c=-\frac{1}{2}$

Therefore, they are parallel.

(B) In order make that the System has Infinite solutions, the slopes and the y-intercepts of both equations must be the same. Then, the values of "a" and "b" can be:

$a=6\\\\b=-4$

If you substitute those values into the second equation and then you solve for "y", you get:

$6x+(-4y)=-2\\\\-4y=-6x-2\\\\y=\frac{-6}{-4}x-\frac{2}{-4}\\\\y=\frac{3}{2}x+\frac{1}{2}$

You can identify that:

$m=\frac{3}{2}\\\\b=\frac{1}{2}$

Therefore, they are the same line.

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3 years ago