Question

3x^2+bx+24=0 if one of its root is 3

Answer 1

Answer:

Other root of equation, $\frac{8}{3}$

Value of ($b$), $-17$

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,

$y=ax^2+bx+c$

To solve this problem, one could use Vieta's theorems. Vieta's theorems state the following, let ($x_1$) and ($x_2$) represent the roots of the equation,

$(x_1)+(x_2)=-\frac{b}{a}$

$(x_1)(x_2)=\frac{c}{a}$

Substitute the given values into the equation,

$3+(x_2)=-\frac{b}{3}$

$(3)(x_2)=\frac{24}{3}$

Simplify,

$(3)+(x_2)=-\frac{b}{3}$

$(3)(x_2)=8$

Inverse operations,

$(3)+(x_2)=-\frac{b}{3}$

$x_2=\frac{8}{3}$

Substitute in the value of ($x_2$),

$(3)+(\frac{8}{3})=-\frac{b}{3}$

Simplify, convert to improper fractions and combine like terms,

$\frac{9}{3}+\frac{8}{3}=-\frac{b}{3}$

$\frac{17}{3}=-\frac{b}{3}$

Multiply both sides of the equation by ($3$) to remove the denominator,

$17=-b$