Question

This is for 8th grade pls answer .​

Answer 1

Step-by-step explanation:

We have that

$(x + \frac{1}{x} ) {}^{2} = 3$

We are trying to find the number value so that we can apply in the later equation.

Qe first simplify.

Remeber that

$(a + b) {}^{2} = a {}^{2} + 2ab + {b}^{2}$

Also remeber that

$\frac{1}{x} = {x}^{ - 1}$

so

$(x + x {}^{ - 1} ) {}^{2} = {x}^{2} + 2x {}^{0} + {x}^{ - 2} = 3$

We then simply remeber that x^0=1 so

${x}^{2} + 2 + \frac{1}{ {x}^{2} } = 3$

Multiply both sides by x^2.

${x}^{4} + 2 {x}^{2} + 1 = 3 {x}^{2}$

Subtract both sides by 3x^2

${x}^{4} - {x}^{2} + 1 = 0$

Notice that x^4= (x^2)^2.

So our reformed equation is

$( {x}^{2} ) {}^{2} - {x}^{2} + 1 = 0$

Let a variable , w equal x^2. This means that we subsitute variable, w in for x^2.

$w {}^{2} - w + 1 = 0$

Now we use the quadratic formula

$w = \frac{ - b + \sqrt{b {}^{2} - 4ac } }{2a}$

and

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

Let a=1 b=-1 and c=1.

$w = \frac{1 + \sqrt{1 - 4(1)} }{2}$

$w = \frac{1 - \sqrt{1 - 4} }{2}$

Now, we get

$w = \frac{1}{2} + \frac{i \sqrt{3} }{2}$

and

$w = \frac{1}{2} - \frac{ i\sqrt{3} }{2}$

Now since we set both of these to the x^2 we solve for x.

and

${x}^{2} = \frac{1}{2} + \frac{i \sqrt{3} }{2}$

and

${x}^{2} = \frac{1}{2} - \frac{i \sqrt{3} }{2}$

We can represent both of these as complex number in the form of a+bi. Next we can convert this into trig form which is

${x}^{2} = 1( \cos(60) + i \: \sin(60)$

and

${x}^{2} = 1( \cos(300) + i \: sin(300))$

Next we take the sqr root of 1 which is 1, and divide the degree by two.

${x} = 1( \cos(30) + i \: sin \: 30)$

and

$x = 1( \cos(150) + i \: sin(150)$

We are asked for the 2nd root so just add 180 degrees to this and we have

$x = 1 \cos(210) + i \: sin \: 210)$

and

$x = 1( \cos(330) + i \: sin(330)$

which both simplified to

$x = - \frac{ \sqrt{3} }{2} - \frac{1}{2} i$

and

$x = \frac{ \sqrt{3} }{2} - \frac{1}{2} i$

Now we must find

x^18+x^12+x^6+1.

We just use demovire Theorem. Which is a complex number raised to the nth root is

${r}^{n} (cos(nx) + i \: sin(nx)$

So let plug in our first root.

$1( \cos(330 \times 18)) + i \: sin \: (330 \times 18))) + 1( \cos(12 \times 330)) + i \: sin(12 \times 330) + 1( \cos(6 \times 330) + i \: sin(6 \times 330))) + 1$

To save time we multiply the angle and use rules of terminals angle and we get

$1( \cos(180) + i \sin(180) ) + 1( \cos(0) + i \: sin \:( 0) + 1( \cos(180) + i \: sin(180) + 1$

And we get

$- 1 + 1 + - 1 + 1 = 0$

So one of the answer is x=0

And the other, let see

$1 \cos(210 \times 18) + i \: \sin(210 \times 18) + 1 \: cos(210 \times 12) + i \: sin(210 \times 12) + 1 \cos(210 \times 6) + \:i sin(210 \times 6) + 1$

$\cos(180) + i \: sin(180) + 1 \cos(0) + i\sin(0) +1( \cos(0) + i \sin(0) + 1$

We get

$- 1 + 1 + 1 + 1 = 2$

So our answer are 2.

So the answer to the second part is

0 and 2.