Question 887630: Find the value of k that makes the sum of the roots of 3x^2+18=(3k+2)x equal to 6.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 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.
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