Question

find the sum of the product of the roots of the quadratic equation.
x^2-3x+12=0

Answer 1

think you mean the sum AND the product of the roots?

Answer 2

yes sorry

Answer 3

if a quadratic equation has roots r1 and r2, then it can be written as:\[(x-r1)(x-r2)=0\]\[x^2-2(r1+r2)x+r1r2=0\]compare this with your equation to find the sum and the product of the roots.

Answer 4

product of the roots is c/a in the general quadratic ax^2 +bx + c

sum of roots = -b/a product = c/a

Answer 5

the product is 12

Answer 6

so the sum is 3

Answer 7

sorry - my expansion was wrong, should have been:\[(x-r1)(x-r2)=x^2-(r1+r2)x+r1r2=0\]

Answer 8

do you see how to find the answer now?

Answer 9

good work XDDDDD

Answer 10

no where do i get r1 and r2

Answer 11

just use the coefficients of the quadratic... a = 1, b = -3 and c = 12... do worry about factorising...

just use sum = -b/a and product = c/a.... it always works... and is easier

Answer 12

we started by saying that if a quadratic equation has roots r1 and r2, then it can be written as above, which simplified to:\[x^2-(r1+r2)x+r1r2=0\]in your case, your equation is:\[x^2-3x+12=0\]so you just need to compare the coefficients of \(x^2\), \(x\) and the constant term.
that gives you:\[r1+r2=3\]\[r1r2=12\]

Answer 13

so now you know the sum of the roots \(r1+r2\) and their product \(r1r2\).

Answer 14

That was way easier Thanks

Answer 15

that solution only works for a monic quadratic...

Answer 16

I think we need to build up knowledge slowly rather than flooding new comers with generalisations.

Answer 17

I agree that helped me a lot thanks

Answer 18

yw

Answer 19

lol... oh well... hope he doesn't get 2x^2 + 5x - 13 as a question