Respuesta :
7x^2-3x-9=0
applying the quadratic formula we get:
x=[-b+/-sqrt(b^2-4ac)]/(2a)
x=[-(-3)+/-sqrt((-3)^2-4*7*(-9))]/(2*7)
x=[3+/-sqrt(9+252)]/14
x=[3+/-sqrt162]/14
hence
x=[3+/-16.1555]/14
x=1.3682 or -0.9396
thus the sum of the roots will be:
(1.3682+(-0.9396))=0.4286
the product will be:
(1.3682)(-0.9396)
=-1.28556
applying the quadratic formula we get:
x=[-b+/-sqrt(b^2-4ac)]/(2a)
x=[-(-3)+/-sqrt((-3)^2-4*7*(-9))]/(2*7)
x=[3+/-sqrt(9+252)]/14
x=[3+/-sqrt162]/14
hence
x=[3+/-16.1555]/14
x=1.3682 or -0.9396
thus the sum of the roots will be:
(1.3682+(-0.9396))=0.4286
the product will be:
(1.3682)(-0.9396)
=-1.28556
Answer:
Sum of the roots is [tex]\frac{3}{7}[/tex]
Product of the roots is [tex]-\frac{9}{7}[/tex]
Step-by-step explanation:
In a quadratic equation ax² + bx + c = 0,
Sum of the roots = [tex]-\frac{b}{a}[/tex]
And, product of the roots = [tex]\frac{c}{a}[/tex]
Here, the given quadratic equation,
[tex]7x^2-3x-9=0[/tex]
Hence,
Sum of the roots = [tex]-\frac{-3}{7}[/tex] = [tex]\frac{3}{7}[/tex]
Product of the roots = [tex]\frac{-9}{7}[/tex] = [tex]-\frac{9}{7}[/tex]