Question 887630
the sum of the roots are equal to -b/a
the product of the roots are equal to c/a
the standard form of the quadratic equation is ax^2 + bx + c = 0


your equation is:
3x^2 + 18 = (3k+2)x


subtract (3k+2)x from both sides of the equation to get:
3x^2 - (3k+2)x + 18 = 0


this equation is now in standard form of ax^2 + bx + c where:
a = 3
b = -(3k+2)
c = 18


sum of the roots are -b/a which becomes (3k+2)/3.


since sum of the roots is equal to 6, this means that (3k+2)/3 = 6.


multiply both sides of this equation by 3 to get:
3k+2 = 18


subtract 2 from both sides of this equation to get:
3k = 16


divide both sides of this equation by 3 to get;
k = 16/3


that should be the value of k.


let's see if that works.


when k = 16/3, 3k+2 = 3*16/3 + 2 = 18.


your equation of 3x^2 + 18 = (3k+2)x becomes:
3x^2 + 18 = 18x


subtract 18x from both sides of this equation to get:
3x^2 - 18x + 18 = 0


divide both sides of this equation by 3 to get:
x^2 - 6x + 6 = 0


we can use the formula of sum of the roots = -b/a to get 6/1 = 6.


let's find the roots to see if the formula gave us the right answer.


the roots of this equation are:


x = (6 - sqrt(12))/ 2 and x = (6 + sqrt(12)) / 2


the sum of the roots are (6 - sqrt(12)) / 2 + (6 + sqrt(12)) / 2


this becomes:


sum = (6 - sqrt(12) + 6 + sqrt(12)) / 2


simplify this to get sum = 12 / 2 because the + sqrt(12) and the - sqrt(12) cancel out.


simplify further to get sum = 6.


the formula for sum of roots = -b/a works and the solution is correct.