2017年7月24日 · See below x=1.62 or x= -0.62 The quadratic formula is [-b+-sqrt(b^2-4ac)]/(2a) this is used for formulas ax^2+bx+c=0 in your problem, " "x^2−x−1=0" " a=1, b=-1, c ...

2015年4月11日 · You can use the standard formula which allows you to solve any quadratic equation, which is x_{1,2} = \frac{-b \pm \sqrt{b^2-4ac}}{2a} b^2-4ac, also called \Delta (delta), is the discriminant: if \Delta\ge 0, then the solutions are real numbers (you have two distinct solutions if \Delta > 0 and two solutions collapsed into the same point if \Delta=0), while if \Delta<0, the two solutions are ...

2016年6月28日 · x - arctan x + C x^2/(x^2+1) = (x^2+1 - 1)/(x^2+1) = 1 - ( 1)/(x^2+1) int \ 1 - ( 1)/(x^2+1) \ dx = x - color(red)(int \ ( 1)/(x^2+1) \ dx ) in terms of the red bit, use sub x = tan t, dx = sec^2 t \ dt this makes it \int \ ( 1)/(tan^2 t+1) \ sec^2 t \ dt = \int \ ( 1)/(sec^2 t) \ sec^2 t \ dt = \int \ dt = arctan x - C So the full integral is ...

2016年11月21日 · How do you use Integration by Substitution to find #intx/(x^2+1)dx#? How do you use Integration by Substitution to find #inte^x*cos(e^x)dx#? See all questions in Integration by Substitution

2015年5月4日 · To factor #x^2+x+1# use the quadratic formula for roots #x=(-b+-sqrt(b^2-4ac))/(2a)# In this case: #x= (-1+-sqrt(1^2-4(1)(1)))/(2(1))# #x = -1/2 +-sqrt(-3)/2# With no Real roots and thus no factors if we are restricted to Real numbers. However, if we are allowed to venture into Complex numbers #x= -1/2 +-sqrt(3)/2i# and the factors are

2016年11月25日 · lim_(x rarr 1)(x^2-1)/(x-1) = 2 Let f(x) = (x^2-1)/(x-1) then f(x) is defined everywhere except at x=1, however when we evaluate the limit we are not interested in ...

2017年10月19日 · The graph is of general shape #uu# as the #x^2# term is positive Given: #y=ax^2+bx+c color(white)("d")->color(white)("d")x^2+x+1#

2017年3月2日 · How do you use the Binomial Theorem to expand #(1 + x) ^ -1#? Precalculus The Binomial Theorem The Binomial Theorem. 1 Answer

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