Respuesta :
Answer:
Part 1) The sum of the roots is [tex]4/3[/tex]
Part 2) The product of the roots is [tex]-7/3[/tex]
Step-by-step explanation:
Step 1
Find the roots
we have
[tex]3x^{2}-4x-7=0[/tex]
The formula to solve a quadratic equation of the form [tex]ax^{2} +bx+c=0[/tex] is equal to
[tex]x=\frac{-b(+/-)\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]3x^{2}-4x-7=0[/tex]
so
[tex]a=3\\b=-4\\c=-7[/tex]
substitute in the formula
[tex]x=\frac{-(-4)(+/-)\sqrt{-4^{2}-4(3)(-7)}} {2(3)}[/tex]
[tex]x=\frac{4(+/-)\sqrt{16^{2}+84}} {6}[/tex]
[tex]x=\frac{4(+/-)10} {6}[/tex]
[tex]x1=\frac{4+10} {6}=7/3[/tex]
[tex]x2=\frac{4-10} {6}=-1[/tex]
Step 2
Find the sum of the roots
[tex]x1+x2=(7/3)-1=4/3[/tex]
Step 3
Find the product of the roots
[tex]x1*x2=(7/3)*(-1)=-7/3[/tex]