Use calculus to show that x + 6x that f(x) = x³ + 6x Briefly, why does part (a) show that a = 2? Note that 10 + V108 = 10 + 6/3. Suppose that we know that the cube-root of 10 + V108 has the form c+ d/3 where c, d are integers. What are the values of c and d? (Hint: Compute the cube of c+ dV3.) Use a method similar to part (c) to compute the cube-root of -10 + V108. Compute a by subtracting the result of part (d) from the result of part (c). What do you obtain? = 20 has a unique real solution. (Hint: Show - 20 has a real root and that it must be unique.)
Use calculus to show that x + 6x that f(x) = x³ + 6x Briefly, why does part (a) show that a = 2? Note that 10 + V108 = 10 + 6/3. Suppose that we know that the cube-root of 10 + V108 has the form c+ d/3 where c, d are integers. What are the values of c and d? (Hint: Compute the cube of c+ dV3.) Use a method similar to part (c) to compute the cube-root of -10 + V108. Compute a by subtracting the result of part (d) from the result of part (c). What do you obtain? = 20 has a unique real solution. (Hint: Show - 20 has a real root and that it must be unique.)
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![1. Cardano’s cubic formula gives \( \alpha = \sqrt[3]{10 + \sqrt{108}} - \sqrt[3]{-10 + \sqrt{108}} \) as a solution to \( x^3 + 6x = 20 \). It is quickly checked that 2 is also a solution. A calculator suggests that \( \alpha = 2 \). In this problem we give two separate verifications that \( \alpha = 2 \).
(a) Use calculus to show that \( x^3 + 6x = 20 \) has a unique real solution. (Hint: Show that \( f(x) = x^3 + 6x - 20 \) has a real root and that it must be unique.)
(b) Briefly, why does part (a) show that \( \alpha = 2 \)?
(c) Note that \( 10 + \sqrt{108} = 10 + 6\sqrt{3} \). Suppose that we know that the cube-root of \( 10 + \sqrt{108} \) has the form \( c + d\sqrt{3} \) where \( c, d \) are integers. What are the values of \( c \) and \( d \)? (Hint: Compute the cube of \( c + d\sqrt{3} \).)
(d) Use a method similar to part (c) to compute the cube-root of \( -10 + \sqrt{108} \).
(e) Compute \( \alpha \) by subtracting the result of part (d) from the result of part (c). What do you obtain?
2. In the 19th century it was finally shown that polynomial equations of degree 5 and higher need not be “solvable by radicals.”
(a) Give an example of a 5th degree polynomial equation that *is* solvable by radicals and give its real solution(s).
(b) The equation \( x^5 - 6x + 3 = 0 \) is not solvable by radicals. Sketch the important features of the graph of \( f(x) = x^5 - 6x + 3 \) and give approximate values of its real roots.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F04075af6-6751-4482-8059-443336c7a451%2F8c427890-f2a5-4e67-ad83-c655c58224ac%2Ftwn2zfc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:1. Cardano’s cubic formula gives \( \alpha = \sqrt[3]{10 + \sqrt{108}} - \sqrt[3]{-10 + \sqrt{108}} \) as a solution to \( x^3 + 6x = 20 \). It is quickly checked that 2 is also a solution. A calculator suggests that \( \alpha = 2 \). In this problem we give two separate verifications that \( \alpha = 2 \).
(a) Use calculus to show that \( x^3 + 6x = 20 \) has a unique real solution. (Hint: Show that \( f(x) = x^3 + 6x - 20 \) has a real root and that it must be unique.)
(b) Briefly, why does part (a) show that \( \alpha = 2 \)?
(c) Note that \( 10 + \sqrt{108} = 10 + 6\sqrt{3} \). Suppose that we know that the cube-root of \( 10 + \sqrt{108} \) has the form \( c + d\sqrt{3} \) where \( c, d \) are integers. What are the values of \( c \) and \( d \)? (Hint: Compute the cube of \( c + d\sqrt{3} \).)
(d) Use a method similar to part (c) to compute the cube-root of \( -10 + \sqrt{108} \).
(e) Compute \( \alpha \) by subtracting the result of part (d) from the result of part (c). What do you obtain?
2. In the 19th century it was finally shown that polynomial equations of degree 5 and higher need not be “solvable by radicals.”
(a) Give an example of a 5th degree polynomial equation that *is* solvable by radicals and give its real solution(s).
(b) The equation \( x^5 - 6x + 3 = 0 \) is not solvable by radicals. Sketch the important features of the graph of \( f(x) = x^5 - 6x + 3 \) and give approximate values of its real roots.
Expert Solution
Solution b:
From part a, the solution is real and unique.
Also, there is no local maximum or minimum.
Thus, the function is increasing.
Therefore it will intersect at only one real value of x.
And it is given that 2 is the solution of the function.
Hence,
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