
What is the quadratic equation of 17x^2 = 12x? | Socratic
2015年10月21日 · 17x^2-12x=0 The general form of the quadratic equation is: ax^2+bx+c=0 in this case we have: 17x^2=12x=> subtract 12x from both sides: 17x^2-12x=0=> in general form …
What is x in x^2 = 17? - Socratic
2015年12月27日 · x = +-sqrt(17) ~~ +-4.123 The equation x^2 = 17 has two solutions, which are the square roots of 17. The principal square root of 17 is the positive one, which is what we …
How do you solve x^2 + 17 = 0? | Socratic
2015年11月2日 · No real roots This can be written as x^2 = -17 x = +-sqrt(-17) Well you have 2 possible complex root but they are real
How do you solve using the quadratic formula #17x^2 = 12x
#17x^2-12x=0# this is now in the form: #ax^2+bx+c=0# With: #a=17# #b=-12# #c=0# That can be used in the quadratic formula to get your two Real solutions (if possible) for #x# as: #x_(1,2)=( …
How do you long divide #(x^2+x - 17) / (x-4)#? - Socratic
2016年4月5日 · x+5, remainder = 3 Look at the first terms of the divisor and the dividend, x and x^2. When x is multiplied by x, we get x^2. This x is the first term of the quotient. This x is …
How do you solve the quadratic equation by completing the …
2015年7月14日 · x=3+sqrt(17), 3-sqrt(17) x^2-6x-8=0 In order to solve a quadratic equation by completing the square, we must force a perfect square trinomial on the left side. …
How do you integrate #1/ (x^2 -2x + 17)#? - Socratic
2015年4月12日 · Answer #int 1/(x^2 - 2x + 17)"dx" = 1/4tan^-1((x - 1)/4) + C#. Method The idea behind this one is: you want to make that base( denominator: # x^2 - 2x + 17 #) look …
How do you factor x^2 + 17x + 42 =0? - Socratic
-3, -14 x^2 + 17x +42 = 0 or, x^2 + 14x + 3x +42 = 0 or, x(x + 14) + 3 (x + 14) = 0 or, (x + 3)(x+14) = 0 or, x= -3, -14
How do you solve x^2-10x+8=0 by completing the square?
2015年4月29日 · Write it as: x^2-10x=-8 add and subtract 25; x^2-10x+25-25=-8 (x-5)^2-25=-8 (x-5)^2=17 x-5=+-sqrt(17) x_(1,2)=5+-sqrt(17)
How do you solve the system x^2+y^2=17 and y=x+3? - Socratic
2016年8月30日 · The Soln. Set ={(1,4),(-4,-1)}. Sub.ing y=x+3 in the first eqn. to get. x^2+(x+3)^2=17, or, 2x^2+6x+9-17=0, i.e., x^2+3x-4=0 rArr (x-1)(x+4)=0 rArr x=1, x=-4.