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Tomtit [17]
3 years ago
10

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

Mathematics
1 answer:
Likurg_2 [28]3 years ago
8 0

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

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