Pre Cal Prob.?

Question

Find the sum and product of the roots of 6x^3 - 9x^2 + x = 0

Please help by showing work.

Answer 1

Since you're in precal, it's possible you don't know this. However, given any polynomial (i'll take a cubic since that's what you have):

ax^3+bx^2+cx+d=0

the SUM of the roots is -b/a
the SUM of the PRODUCT of the roots TWO at a time is c/a
the PRODUCT of the roots is -d/a

The proof of the above is not hard, but tell me if you want to know why it's true.
The answer to your specific problem is sum=9/6=3/2, product=0

Answer 2

Well, you've got to factor it.

x could be zero

x * 6x2-9x+1 = 0

Running through the quadratic formula...

( 9 (+/-) (81 - 24)^.5) /12

Now, that horrible radical term cancels itself out in the addition - remember, both plus and minus. So we worry about the the lead term 9/12, or 3/4. The sum is 6/4, or 1.5

The product is zero, since one of the roots is zero.

Answer 3

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Answer 4

6x^3 - 9x^2 + x = 0
x(6x^2 - 9x + 1) = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(-9) ± sqrt((-9)^2 - 4(6)(1)))/(2(6))
x = (9 ± sqrt(81 - 24))/12
x = (9 ± sqrt(57))/12
x = (1/12)(9 ± sqrt(57))

since one of the factors is 0, the product of the roots is zero.

(1/12)(9 + sqrt(57)) + (1/12)(9 - sqrt(57))
(1/12)((9 + sqrt(57)) + (9 - sqrt(57)))
(1/12)(9 + sqrt(57) + 9 - sqrt(57))
(1/12)(9 + 9)
(1/12)(18)
(18/12)
(3/2)

The Sum of the roots is (3/2)

Answer 5

the product of roots is constant/ coefficent of x^3 =0

sum of roots = -coffecient o x^2/ coefficint of x3 = 9/6

proof:
let roots be a , b , c
then (x-a)(x-b)(x-c) =0

the constant term = abc
coewfficient of x^2 = -(a+b+c)

if coefficient of x^3 is d we need to multiply by d to get equation

Answer 6

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