SOLUTION: I got two questions the first is a quadratic expression: 9k^2+66k+21 Next is a quadratic equation: 3k^2-39k=-120

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equation Customizable Word Problems -> SOLUTION: I got two questions the first is a quadratic expression: 9k^2+66k+21 Next is a quadratic equation: 3k^2-39k=-120      Log On


   

Question 775824: I got two questions the first is a quadratic expression:
9k^2+66k+21
Next is a quadratic equation:
3k^2-39k=-120
Answer by tanjo3(60) About Me  (Show Source):
You can put this solution on YOUR website!
(1) Factor out a 3 from the expression.
3(3k^2 + 22k + 7)

(2) Observe that the expression is in the form ak^2 + bk + c. Here we can apply the quadratic forumla.
k+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
k+=+%28-22+%2B-+sqrt%28+22%5E2-4%2A3%2A7+%29%29%2F%282%2A3%29+
k+=+%28-22+%2B-+sqrt%28+400+%29%29%2F6+
k+=+%28-22+%2B-+20%29%2F6+
k = -1%2F3, -7
(3) Place you zeros back into the expression.
3[(3k + 1)(k + 7)]

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(1) Move all terms to one side.
3k^2 - 39k + 120 = 0
(2) Factor out a 3 from the expression.
3(k^2 - 13k + 40) = 0

(3) Observe that the expression is in the form ak^2 + bk + c. Find two number which add up b and multiply to c.
-5 + -8 = 13; -5 × -8 = 40
(4) Substitute these numbers in 3[(k + ___) (k + ___)]
3[(k - 5) (k - 8)]