Help please!
1. What are the real or imaginary solutions of the polynomial equation?
x^4-52x^2+576=0
A) 4,-4
B) 4,-6
C) 4,-4,6,-6
D) no solutions
2. What are the real or imaginary solutions of the polynomial equation?
x^3=216
A) -6,3+3i(sq root of 7), and 3-3i(sq root of 7)
B) -6, 3+3i(sq root of 3), and 3-3i(sq root of 3)
C) 6, 3+3i(sq root of 7), and 3-3i(sq root of 7)
D) 6, 3+3i(sq root of 3), and 3-3i(sq root of 3)
5. The dimensions, in inches, of a shipping box at W Ship 4 You can be expressed as width x, length x+5, and height 3x-1. The volume is about 7.6 ft^3. Find the dimension of the box in inches. Round to the nearest inch. (Hint: 1 ft^3=1,728 in.^3)
A) 15 in. by 20 in. by 44 in.
B) 12 in. by 17 in. by 35 in.
C) 15 in. by 20 in. by 45 in.
D) 12 in. by 17 in. by 36 in.
1. To find the solutions of the polynomial equation x^4 - 52x^2 + 576 = 0, we can use substitution or factoring. Let's use factoring:
x^4 - 52x^2 + 576 = (x^2 - 36)(x^2 - 16) = 0
Now, set each factor equal to zero:
x^2 - 36 = 0 or x^2 - 16 = 0
Solving the first equation, x^2 - 36 = 0, we can factor it further:
(x - 6)(x + 6) = 0
Setting each factor equal to zero:
x - 6 = 0, x + 6 = 0
Solving for x gives us two real solutions: x = 6 and x = -6.
Solving the second equation, x^2 - 16 = 0, we can factor it further:
(x - 4)(x + 4) = 0
Setting each factor equal to zero:
x - 4 = 0, x + 4 = 0
Solving for x gives us two additional real solutions: x = 4 and x = -4.
Therefore, the real solutions of the polynomial equation x^4 - 52x^2 + 576 = 0 are 4, -4, 6, and -6.
Answer: C) 4, -4, 6, -6
2. To find the solutions of the polynomial equation x^3 = 216, we can use the cube root:
x = ∛216
The cube root of 216 is 6, since 6 x 6 x 6 = 216.
Therefore, x = 6 is one real solution.
Now, let's find the imaginary solutions. We can rewrite the equation as:
(x - 6)(x^2 + 6x + 36) = 0
Setting each factor equal to zero:
x - 6 = 0
x^2 + 6x + 36 = 0
Solving the first equation, x - 6 = 0, gives us x = 6 as the real solution (which we already found).
Now, let's solve the second equation, x^2 + 6x + 36 = 0, using the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 6, and c = 36. Plugging these values into the quadratic formula, we get:
x = (-6 ± √(6^2 - 4(1)(36))) / (2(1))
x = (-6 ± √(36 - 144))/2
x = (-6 ± √(-108))/2
x = (-6 ± √(-36 * 3))/2
x = (-6 ± 6√3i)/2
x = -3 ± 3√3i
Therefore, the real or imaginary solutions of the polynomial equation x^3 = 216 are: -6, -3 + 3√3i, and -3 - 3√3i.
Answer: B) -6, 3 + 3√3i, and 3 - 3√3i.
3. To find the dimensions of the shipping box in inches, we need to solve for x in the equation:
(width)(length)(height) = 7.6 ft^3
Since 1 ft^3 = 1,728 in^3, we can convert 7.6 ft^3 to inches as follows:
7.6 ft^3 * 1,728 in^3/ft^3 = 13,123.2 in^3
Now, substitute the expressions for the dimensions in terms of x:
(x)(x + 5)(3x - 1) = 13,123.2
Expand the equation:
3x^3 + 11x^2 - 5x - 13,123.2 = 0
Unfortunately, this is a cubic equation, and solving it directly can be complex.
However, given the answer choices, we can iterate through the dimensions provided and calculate their volume in inches using the formula:
Volume = (width)(length)(height)
Checking each answer choice:
A) (15)(20)(44) = 13,200 in^3
B) (12)(17)(35) = 7,140 in^3
C) (15)(20)(45) = 13,500 in^3
D) (12)(17)(36) = 8,208 in^3
The closest answer choice to 13,123.2 in^3 is B) 12 in. by 17 in. by 35 in.
Answer: B) 12 in. by 17 in. by 35 in.
1. To find the real or imaginary solutions of the polynomial equation x^4-52x^2+576=0, we can use the quadratic equation formula. Let's rewrite the equation in terms of a quadratic equation by substituting x^2 as a new variable. Let's say y = x^2. Now the equation becomes y^2 - 52y + 576 = 0. We can solve this quadratic equation using the quadratic formula, which states that for any quadratic equation of the form ax^2 + bx + c = 0, the solutions are given by x = (-b ± √(b^2 - 4ac))/(2a).
Applying this formula to our quadratic equation y^2 - 52y + 576 = 0, we have a = 1, b = -52, and c = 576. Plugging these values into the formula, we get y = (52 ± √(52^2 - 4*1*576))/(2*1). Simplifying further, we have y = (52 ± √(2704 - 2304))/2. Continuing to simplify, y = (52 ± √400)/2. This gives us two possible solutions: y = (52 + √400)/2 = 26 and y = (52 - √400)/2 = 26.
Since we defined y as x^2, we can solve for x by taking the square root of each solution. Taking the square root of 26, we get two values: x = ±√26. Therefore, the real or imaginary solutions of the polynomial equation x^4-52x^2+576=0 are x = √26 and x = -√26. However, these solutions are not among the choices A), B), C), or D). Therefore, the correct answer is D) no solutions.
2. To find the real or imaginary solutions of the polynomial equation x^3 = 216, we need to solve for x. Taking the cube root of both sides of the equation, we get x = ∛216. The cube root of 216 is 6 since 6^3 = 216. Therefore, the only real solution to the equation x^3 = 216 is x = 6.
Now let's check the answer options to see if any of them match. Option A) includes complex numbers, which means it is not a real solution. Option B) includes a complex number as well. Option C) includes the real solution x = 6 but also includes complex numbers. Option D) only includes real solutions, so it is the correct answer. Therefore, the real or imaginary solutions of the polynomial equation x^3 = 216 are x = 6, 3 + 3i√3, and 3 - 3i√3, which corresponds to answer option D).
3. To find the dimensions of the shipping box in inches, we can use the volume formula for a rectangular prism, which is V = lwh, where V is the volume and l, w, h are the length, width, and height, respectively.
Given that the volume is 7.6 ft^3, we need to convert it to cubic inches since the dimensions are in inches. Using the hint provided, we know that 1 ft^3 = 1728 in^3. So, 7.6 ft^3 is equal to 7.6 * 1728 in^3 = 13132.8 in^3.
Now we can set up the equation using the given dimensions:
x * (x+5) * (3x-1) = 13132.8
Expanding and simplifying the equation, we get:
3x^3 - 2x^2 + 5x - 13132.8 = 0
To solve this cubic equation, we can use numerical methods or calculators. After solving the equation, the approximate value for x is 12.
Now we can substitute the value of x back into the dimensions. The width is x inches, so it becomes 12 inches; the length is x+5 inches, so it becomes 17 inches (12+5); and the height is 3x-1 inches, so it becomes 35 inches (3*12-1).
Therefore, the dimensions of the shipping box in inches are 12 inches by 17 inches by 35 inches, which corresponds to answer option B).
#1. x^4-52x^2+576 = 0
Easy enough to check. They even give you the values to try. Just do a few synthetic divisions to see whether the proposed values work. Synthetic division is fast and easy. Once you have found two roots, you end up with a quadratic, and that's no trouble.
However, I'd try C first, since you can just divide by x^2-16 or x^2-36 to check it.
Or, if you let u = x^2, you have a quadratic in u:
u^2 - 52u + 576 = (u-16)(u-36)
That means the factors are (x^2-16)(x^2-36)
#2. x^3 = 216 = 6^3
So, you know the real root is 6
The complex roots are 6cis(2π/3) and 6cis(4π/3)
or, 6(1±i√3)/2
#3. More of the same. You want to solve
x(x+5)(3x-1) = 7.6*1728
3x^3+14x^2-5x = 13132.8
3x^3 + 14x^2 - 5x - 13133 = 0
Now, you can go through the motions of trying to solve this, but a few seconds with your calculator shows that A gives a volume of 13200, which is pretty close. C and D are obviously wrong, since the dimensions do not fit the requirements.