#1 b-e 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.

Name: #1 b-e 1. Cardano’s cubic formula gives \( \alpha = \sqrt[3]{10 + \sqr
Uploaded: 2024-03-20T07:06:41.000Z
Channel: Bartleby
Description: #1 b-e 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