What are the zero(s) of: #3x^2 - 5x -4 = 0#?
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"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
May 22, 2018
Explanation:
Given:
#3x^2-5x-4 = 0#
The difference of squares identity can be written:
#A^2-B^2 = (A-B)(A+B)#
Complete the square and use this with
#0 = 12(3x^2-5x-4)#
#color(white)(0) = 36x^2-60x-48#
#color(white)(0) = (6x)^2-2(6x)(5)+25-73#
#color(white)(0) = (6x-5)^2-(sqrt(73))^2#
#color(white)(0) = ((6x-5)-sqrt(73))((6x-5)+sqrt(73))#
#color(white)(0) = (6x-5-sqrt(73))(6x-5+sqrt(73))#
Hence:
#6x = 5+-sqrt(73)#
Hence:
#x = 5/6+-sqrt(73)/6#
