Question

Find the zeroes of p(x) 3x^2+13x-10 an verify the relationship between the zeroes and the coefficients. Show working as well.

Answer 1

sum of zeroes= -b/a = -13/3

product of zeroes =c/a= -10/3

Answer 2

Answer:

Step-by-step explanation:

$3x^{2} +13x-10=0$

$3x^{2} +15x-2x-10=0\\3x(x+5)-2(x+5)=0\\(3x-2)(x+5)=0\\x=\frac{2}{3}\\ or\\x=-5$

sum of zeroes=$\frac{-b}{a}$

$\alpha +\beta =\frac{-b}{a}\\\frac{2}{3} +(-5)=\frac{-13}{3}$

take L.C.M.

$\frac{2-15}{3} =\frac{-13}{3}\\\frac{-13}{3}= \frac{-13}{3}$

product of zeroes=$\frac{c}{a}$

$\alpha \beta =\frac{c}{a}\\\frac{2}{3}*(-5)=\frac{-10}{3}\\ \frac{-10}{3} =\frac{-10}{3}$

hence zeroes are equal to their coefficients.