Find the real or imaginary solutions.

x^4 + 3x^2 - 18 = 0

PLEASE SHOW WORK.

Step-by-step explanation:

Here is your explanation:

((x4) - 3x2) - 18 = 0 Factoring  x4-3x2-18 

The first term is,  x4  its coefficient is  1 .

The middle term is,  -3x2  its coefficient is  -3 .

The last term, "the constant", is  -18 

Step-1 : Multiply the coefficient of the first term by the constant   1 • -18 = -18 

Step-2 : Find two factors of  -18  whose sum equals the coefficient of the middle term, which is   -3 .

     -18   +   1   =   -17     -9   +   2   =   -7     -6   +   3   =   -3   That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -6  and  3 

                     x4 - 6x2 + 3x2 - 18

Step-4 : Add up the first 2 terms, pulling out like factors :

                    x2 • (x2-6)

              Add up the last 2 terms, pulling out common factors :

                    3 • (x2-6)

Step-5 : Add up the four terms of step 4 :

                    (x2+3)  •  (x2-6)

             Which is the desired factorization

Find roots (zeroes) of :       F(x) = x2+3

Polynomial Roots Calculator is a set of methods aimed at finding values of  x  for which   F(x)=0  

Rational Roots Test is one of the above mentioned tools. It would only find Rational Roots that is numbers  x  which can be expressed as the quotient of two integers

The Rational Root Theorem states that if a polynomial zeroes for a rational number  P/Q   then  P  is a factor of the Trailing Constant and  Q  is a factor of the Leading Coefficient

In this case, the Leading Coefficient is  1  and the Trailing Constant is  3.

 The factor(s) are:

of the Leading Coefficient :  1

 of the Trailing Constant :  1 ,3

 Let us test ....

  P  Q  P/Q  F(P/Q)   Divisor     -1     1      -1.00      4.00        -3     1      -3.00      12.00        1     1      1.00      4.00        3     1      3.00      12.00   

Factoring:  x2-6 

Theory : A difference of two perfect squares,  A2 - B2  can be factored into  (A+B) • (A-B)

Proof :  (A+B) • (A-B) =

         A2 - AB + BA - B2 =

         A2 - AB + AB - B2 =

         A2 - B2

Note :  AB = BA is the commutative property of multiplication.

Note :  - AB + AB equals zero and is therefore eliminated from the expression.

Check : 6 is not a square !!(x2 + 3) • (x2 - 6) = 0 product of several terms equals zero. 

 When a product of two or more terms equals zero, then at least one of the terms must be zero. 

 We shall now solve each term = 0 separately 

 In other words, we are going to solve as many equations as there are terms in the product 

 Any solution of term = 0 solves product = 0 as well.Solve  :    x2+3 = 0 

 Subtract  3  from both sides of the equation : 

                      x2 = -3

 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  

                      x  =  ± √ -3  

 In Math,  i  is called the imaginary unit. It satisfies   i2  =-1. Both   i   and   -i   are the square roots of   -1 

Accordingly,  √ -3  =

                    √ -1• 3   =

                    √ -1 •√  3   =

                    i •  √ 3

The equation has no real solutions. It has 2 imaginary, or complex solutions.

                      x=  0.0000 + 1.7321 i 

                      x=  0.0000 - 1.7321 i 

Solve  :    x2-6 = 0 

 Add  6  to both sides of the equation : 

                      x2 = 6

 When two things are equal, their square roots are equal. Taking the square root of the two sides of the equation we get:  

                      x  =  ± √ 6  

 The equation has two real solutions  

 These solutions are  x = ± √6 = ± 2.4495  

 Solving  x4-3x2-18  = 0 directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Solve   x4-3x2-18 = 0

This equation is reducible to quadratic. What this means is that using a new variable, we can rewrite this equation as a quadratic equation Using  w , such that  w = x2  transforms the equation into :

 w2-3w-18 = 0

Solving this new equation using the quadratic formula we get two real solutions :

   6.0000  or  -3.0000

Now that we know the value(s) of  w , we can calculate  x  since  x  is  √ w  

Doing just this we discover that the solutions of

   x4-3x2-18 = 0

  are either : 

  x =√ 6.000 = 2.44949  or :

  x =√ 6.000 = -2.44949  or :

  x =√-3.000 = 0.0 + 1.73205 i  or :

  x =√-3.000 = 0.0 - 1.73205 i

so we found four answers:

 x = ± √6 = ± 2.4495

  x=  0.0000 - 1.7321 i 

  x=  0.0000 + 1.7321 i 

sorry for taking time to answer

hope its helps

The correct option is A because [tex]x=\pm\sqrt{6}i[/tex] or [tex]x=\pm\sqrt{3}[/tex].

Given:

The given equation is:

[tex]x^4+3x^2-18=0[/tex]

To find:

The real and imaginary solutions of the given equation.

Explanation:

We have,

[tex]x^4+3x^2-18=0[/tex]

Substitute [tex]x^2=t[/tex] in the given equation.

[tex]t^2+3t-18=0[/tex]

Splite the middle term.

[tex]t^2+6t-3t-18=0[/tex]

[tex]t(t+6)-3(t+6)=0[/tex]

[tex](t+6)(t-3)=0[/tex]

Using zero product property, we get

[tex]t+6=0[/tex] or [tex]t-3=0[/tex]

[tex]t=-6[/tex] or [tex]t=3[/tex]

Substituting [tex]t=x^2[/tex], we get

[tex]x^2=-6[/tex] or [tex]x^2=3[/tex]

[tex]x=\pm\sqrt{-6}[/tex] or [tex]x=\pm\sqrt{3}[/tex]

[tex]x=\pm\sqrt{6}i[/tex] or [tex]x=\pm\sqrt{3}[/tex]

Therefore, the correct option is A.

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