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In Mathematics / College | 2025-07-07

Solve for $x$ in the equation $3 x^2-18 x+5=47$

A. $x=3 \pm \sqrt{23}$
B. $x=3 \pm \sqrt{51}$
C. $x=3 \pm \sqrt{41}$
D. $x=3 \pm \sqrt{5}$

Asked by gapinskikarla

Answer (1)

Rewrite the equation in standard quadratic form: 3 x 2 − 18 x − 42 = 0 .
Simplify the equation by dividing by 3: x 2 − 6 x − 14 = 0 .
Apply the quadratic formula: x = 2 a − b ± b 2 − 4 a c ​ ​ = 2 ( 1 ) 6 ± ( − 6 ) 2 − 4 ( 1 ) ( − 14 ) ​ ​ .
Simplify to find the solutions: x = 3 ± 23 ​ .

Explanation

Problem Analysis We are given the equation 3 x 2 − 18 x + 5 = 47 . Our goal is to solve for x .

Rewrite the Equation First, we need to rewrite the equation in the standard quadratic form, which is a x 2 + b x + c = 0 . To do this, we subtract 47 from both sides of the equation:

Simplify 3 x 2 − 18 x + 5 − 47 = 0 , which simplifies to 3 x 2 − 18 x − 42 = 0 .

Divide by Constant Now, we can divide the entire equation by 3 to simplify it further:

Simplified Equation x 2 − 6 x − 14 = 0 .

Apply Quadratic Formula We can use the quadratic formula to solve for x . The quadratic formula is given by x = 2 a − b ± b 2 − 4 a c ​ ​ , where a = 1 , b = − 6 , and c = − 14 .

Substitution Substituting the values of a , b , and c into the quadratic formula, we get:

Calculation x = 2 ( 1 ) − ( − 6 ) ± ( − 6 ) 2 − 4 ( 1 ) ( − 14 ) ​ ​

Simplify Simplifying the expression:

Further Simplification x = 2 6 ± 36 + 56 ​ ​

Simplify Square Root x = 2 6 ± 92 ​ ​

Square Root Simplification We can simplify the square root: 92 ​ = 4 ⋅ 23 ​ = 2 23 ​ .

Substitute Back Substituting back into the equation:

Final Simplification x = 2 6 ± 2 23 ​ ​

Divide by 2 Dividing both terms in the numerator by 2, we get:

Final Answer x = 3 ± 23 ​


Examples
Imagine you are designing a rectangular garden where the area is related to the length by a quadratic equation. Solving such an equation helps you determine the possible lengths of the garden given a specific area constraint. Quadratic equations are also used in physics to model projectile motion, helping to calculate the trajectory of objects, like a ball thrown in the air. Understanding how to solve these equations allows for precise calculations in design and physics, ensuring accuracy and efficiency in real-world applications.

Answered by GinnyAnswer | 2025-07-07