Solve for $x$ in the equation $x^2-8 x+41=0$.

A. $x-4 \pm \sqrt{37}$
B. $x=-4 \pm 5$
C. $x=4 \pm \sqrt{37} i$
D. $x=4 \pm 5 i$

To solve the equation x 2 − 8 x + 41 = 0 , we use the quadratic formula and find that the solutions are complex: 4 ± 5 i . Therefore, the answer is option D: x = 4 ± 5 i .
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Answered by Anonymous | 2025-07-04

Use the quadratic formula x = 2 a − b ± b 2 − 4 a c ​ ​ with a = 1 , b = − 8 , and c = 41 .
Substitute the values into the formula: x = 2 8 ± ( − 8 ) 2 − 4 ( 1 ) ( 41 ) ​ ​ .
Simplify the expression: x = 2 8 ± − 100 ​ ​ = 2 8 ± 10 i ​ .
Divide by 2 to get the final answer: x = 4 ± 5 i .

Explanation

Problem Analysis We are given the quadratic equation x 2 − 8 x + 41 = 0 . Our goal is to find the values of x that satisfy this equation.

Applying the Quadratic Formula We can solve this quadratic equation using the quadratic formula, which is given by: x = 2 a − b ± b 2 − 4 a c ​ ​ where a , b , and c are the coefficients of the quadratic equation a x 2 + b x + c = 0 . In our case, a = 1 , b = − 8 , and c = 41 .

Simplifying the Expression Substituting the values of a , b , and c into the quadratic formula, we get: x = 2 ( 1 ) − ( − 8 ) ± ( − 8 ) 2 − 4 ( 1 ) ( 41 ) ​ ​ Simplifying this expression, we have: x = 2 8 ± 64 − 164 ​ ​ x = 2 8 ± − 100 ​ ​

Dealing with Complex Numbers Since the discriminant (the value inside the square root) is negative, we will have complex solutions. We can rewrite − 100 ​ as 100 ​ ⋅ − 1 ​ = 10 i , where i is the imaginary unit ( i 2 = − 1 ). So, our equation becomes: x = 2 8 ± 10 i ​

Final Solutions Now, we divide both the real and imaginary parts by 2: x = 4 ± 5 i Thus, the solutions for x are 4 + 5 i and 4 − 5 i .


Examples
Quadratic equations are used in various fields such as physics, engineering, and economics. For example, in physics, projectile motion can be modeled using quadratic equations to determine the trajectory of an object. In engineering, quadratic equations can be used to design parabolic reflectors for antennas or solar panels. In economics, they can be used to model cost and revenue functions to find the break-even point.

Answered by GinnyAnswer | 2025-07-04