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 where: a=17 , b=-12 and c=0

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 mean when we write sqrt(17). sqrt(17) is an irrational number. That is, it cannot be represented as p/q for some integers p and q. As a result, its decimal expansion does not eventually terminate or …

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

#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)=(-b+-sqrt(b^2-4ac))/(2a)# Try by yourself substituting values (you should get two real values).

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 multiplied by divisor, x-4 and subtracted from the dividend. We are now left with 5x - 17. Follow the same method, as above, to find the next term of the quotient. Since no more x terms …

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. a^2+2ab+b^2=(a+b)^2 Add 8 to both sides of the equation. x^2-6x=8 Divide the coefficient of the x term by 2 and square the result.

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 something like #1/(x^2 + 1)#

-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

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)

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.