Answer:
Other root of equation, [tex]\frac{8}{3}[/tex]
Value of ([tex]b[/tex]), [tex]-17[/tex]
Step-by-step explanation:
The given quadratic equation is in standard form. Standard form is the basic format of representing a quadratic equation, a quadratic equation in standard form would use the following format,
[tex]y=ax^2+bx+c[/tex]
To solve this problem, one could use Vieta's theorems. Vieta's theorems state the following, let ([tex]x_1[/tex]) and ([tex]x_2[/tex]) represent the roots of the equation,
[tex](x_1)+(x_2)=-\frac{b}{a}[/tex]
[tex](x_1)(x_2)=\frac{c}{a}[/tex]
Substitute the given values into the equation,
[tex]3+(x_2)=-\frac{b}{3}[/tex]
[tex](3)(x_2)=\frac{24}{3}[/tex]
Simplify,
[tex](3)+(x_2)=-\frac{b}{3}[/tex]
[tex](3)(x_2)=8[/tex]
Inverse operations,
[tex](3)+(x_2)=-\frac{b}{3}[/tex]
[tex]x_2=\frac{8}{3}[/tex]
Substitute in the value of ([tex]x_2[/tex]),
[tex](3)+(\frac{8}{3})=-\frac{b}{3}[/tex]
Simplify, convert to improper fractions and combine like terms,
[tex]\frac{9}{3}+\frac{8}{3}=-\frac{b}{3}[/tex]
[tex]\frac{17}{3}=-\frac{b}{3}[/tex]
Multiply both sides of the equation by ([tex]3[/tex]) to remove the denominator,
[tex]17=-b[/tex]
