2018年5月24日 · #"the factors of + 12 which sum to + 13 are + 12 and + 1"# #x^2+13x+12=(x+12)(x+1)# Answer link. Jacobi J.

2017年7月7日 · What we can do first is complete the square under the radical to give the equivalent expression. #intsqrt((x+2)^2 + 9) dx#

2015年12月15日 · Here, a = 2, b = 17 and c = 26. x = (-b +- sqrt(b^2 - 4ac))/(2a) = (-17 +- sqrt(17^2 - 4*2*26))/(4) " "= (-17 +- 9)/4 Thus, your second and your third roots are x = -2 and x = - 13/2 ===== 3) Finding positive intervals of the polynomial Now that you have the three roots, you know that the function intercepts the x axis three times, at -13/2, -2 ...

2018年7月17日 · Vertex form of equation is y=13( x+7/13)^2-20 3/13 y=13 x^2+14 x+24 or y=13 (x^2+14/13 x)+24 or y=13 {x^2+14/13 x+(7/13)^2}-49/13+24 or y=13( x+7/13)^2-263/13 or y=13 ...

x^4-13x^2-30 = (x^2-15)(x^2+2) = (x-sqrt(15))(x+sqrt(15))(x^2+2) x^4-13x^2-30 is quadratic in x^2. To factor it, first find a pair of factors of 30 whose difference is 13. Such a pair is 15 and 2. That gives us: x^4-13x^2-30 = (x^2-15)(x^2+2) The first of these quadratic factors can be considered a difference of squares: x^2-15 = x^2 - (sqrt(15))^2 = (x-sqrt(15))(x+sqrt(15)) using the ...

2016年4月26日 · 20x^2+13x+2=(4x+1)(5x+2) Use an AC method: Find a pair of factors of AC = 20*2 = 40 with sum B=13. The pair 8, 5 works in that 8xx5=40 and 8+5 = 13 Use this pair to split the middle term and factor by grouping: 20x^2+13x+2 =20x^2+8x+5x+2 =(20x^2+8x)+(5x+2) =4x(5x+2)+1(5x+2) =(4x+1)(5x+2)

2016年7月4日 · #= x [x^2 + 13x + 42]# Now getting the factors of 42 and seeing if the sum of its factors equals to 13 ( middle value). By getting its factors ( #7*6# = 42) we have come to know that 7 + 6 equals to 13.

2018年3月11日 · How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#?

x^11 Expression = x^13/x^2 Here we will use two of the laws of exponents as follows: (i) 1/a^n = a^-n (ii) a^n xx a^m = a^(n+m) We can rewrite the expression as: Expression = 1/x^2 xx x^13 Apply (i) Expression = x^-2 xx x^13 Apply (ii) Expression = x^(-2+13) = x^11

2017年4月13日 · 169/4 Half the coefficient of x which would be -13/2 and then square it which would be 169/4 which 'c' should be equal to. The perfect square would be (x-13/2)^2