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What are the sum (S) and product (P) of the roots of the equation 3x^2 - 7x + 12 = 0?

 Nov 18, 2014

Best Answer 

 #8
avatar+118710 
+5

 Hi Chris,

The sum of the roots of a polynomial of degree n is   

 

sum of the real roots =coefficient of xn1coefficient of xn

 

but there is a lot more to this Chris.

Have a look  at this

 

http://www.mathsisfun.com/algebra/polynomials-sums-products-roots.html

 Nov 18, 2014
 #1
avatar+130491 
+5

 

 

3x^2 - 7x + 12 = 0    this doesn't factor......using the onsite solver, we have

3×x27×x+12=0{x=(95×i7)6x=(95×i+7)6}{x=(76+1.624465724134i)x=76+1.624465724134i}

The sum of the roots =  (14 / 6) =( 7 / 3 )

The product of the roots is

(1/36)[(7 - (√95)i] [ (7 + (√95)i ] =

(1/36) [49 - 95i2] =

(1/36) [ 49 + 95] =

(1/36)[144] =  4

 

 Nov 18, 2014
 #2
avatar+26397 
+5

What are the sum (S) and product (P) of the roots of the equation 3x^2 - 7x + 12 = 0 ?

\small{\text{ $ \begin{array}{rcl} 3x^2 - 7x + 12 & = & 0 \quad | \quad :3 \\ \\ x^2 - \frac{7}{3}x + \frac{12}{3} & = & 0 \\ \\ x^2 \underbrace{- \frac{7}{3}}_{ = -S } x + \underbrace{ 4 }_{=P }& = & 0 \\\\ $ sum (S) of the roots: $x_1+x_2& = &+ \frac{7}{3} \\\\ $ product (P) of the roots: $ x_1*x_2& = & 4 \end{array} $ }}

 Nov 18, 2014
 #3
avatar+130491 
0

Wow!!.....thanks, heureka.....that's a new one on me !!!....I never recognized it before.....  DOH !!!

 

 Nov 18, 2014
 #4
avatar+118710 
+5

Hi Chris,

What Heureka has done has many implications.  it can be used on polynomials of degree higher than 2.

although I'd need to 'think' with it if I was asked a question with a higher degree.

I think it can be used in probablility too although again I am very rusty   

 Nov 18, 2014
 #5
avatar+33659 
+5

Think of it like this:

 

(x - x1)(x - x2) = 0

 

x2 - (x1 + x2)x +x1x2 = 0

.

 Nov 18, 2014
 #6
avatar+130491 
0

Ah.....that makes it much clearer.....!!!

 

 Nov 18, 2014
 #7
avatar+130491 
0

Melody...just playing around with this in WolframAlpha....it appears that in a polynomial of degree n, that the sum of the roots will always be the coefficient on the xn-1 term, unless the sum of the roots = 0. In that case, that power (naturally) isn't represented in the polynomial. The ending constant will (again, naturally) be the product of the roots !!!

 

 Nov 18, 2014
 #8
avatar+118710 
+5
Best Answer

 Hi Chris,

The sum of the roots of a polynomial of degree n is   

 

sum of the real roots =coefficient of xn1coefficient of xn

 

but there is a lot more to this Chris.

Have a look  at this

 

http://www.mathsisfun.com/algebra/polynomials-sums-products-roots.html

Melody Nov 18, 2014
 #9
avatar+130491 
0

Ah...I see....my guess was wrong....thanks for that link....I'm going to look at that very closely...!!!

 

 Nov 18, 2014

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