Solve for the roots in the equation below. x4 + 3x2 - 4 = 0

First, you need to rewrite the expression into binomial form, so you are working with two terms (as you world with a quadratic):
(x²)²-3(x²)-4=0
Now, you can place the x²s into brackets as the coefficient is now 1:
(x²  )(x²  )
Next, find out two numbers that add together to give you -3 and multiply to give -4 (these are the leftover integers after removing the x²s). These two numbers are -4 and 1.
Place the -4 and 1 into the brackets:
(x²-4)(x²+1)=0
Notice that the x²-4 is a difference of two squares, so can be further factorised into (x+2)(x-2)
This leaves you with a final factorisation of:
(x+2)(x-2)(x²+1)=0
Now we handle each bracket individually to obtain our four solutions for x:
x+2=0
x=-2
x-2=0
x=2
x²+1=0
x²=1
x=±1

put like terms together.orremember that the 45 becomes negative when moved to the other side.orthe minuses cancel out.

a. geometricb. g[h] = (3/4)·g[h-1]; g[1] = 384c. g[h] = 384·(3/4)^(h-1)d. g[12] ≈ 16.2183 gallonse. g[12] ≈ 64.8732 quarts

step-by-step explanation:

part a. the differences are decreasing, so we know the sequence is not arithmetic. the ratios are constant (=3/4), so we know the sequence is geometric.

part b. from part a, we know the common ratio is 3/4, so each term is 3/4 of the one before. the initial term is the number of gallons at h=1.

g[h] = (3/4)·g[h-1]

g[1] = 384 . . . . initial value required for a recursive definition

part c. the general formula for a geometric sequence is

a[n] = a[1]·r^(n-1)

we have a[1] = 384, r = 3/4, and the term number given by h. the formula that follows is

g[h] = 384·(3/4)^(h-1)

part d. filling in the value and doing the arithmetic, we have

g[12] = 384·(3/4)^11 ≈ 384·0.042235136 ≈ 16.2183 . . gallons

part e. each gallon is 4 quarts, so

16.2183 gal × 4 qt/gal = 64.8732 qt