Solve for $x$ in the equation $2 x^2-5 x+1=3$
A. $x=\frac{5}{2} \pm \frac{\sqrt{29}}{2}$
B. $x=\frac{5}{2} \pm \frac{\sqrt{41}}{4}$
C. $x=\frac{5}{4} \pm \frac{\sqrt{29}}{2}$
D. $x=\frac{5}{4} \pm \frac{\sqrt{41}}{4}$
Answer (1)
Rewrite the given equation in standard quadratic form: 2 x 2 − 5 x − 2 = 0 .
Apply the quadratic formula: x = 2 a − b ± b 2 − 4 a c .
Substitute the values a = 2 , b = − 5 , and c = − 2 into the formula and simplify.
The solutions for x are x = 4 5 ± 4 41 , so the final answer is x = 4 5 ± 4 41 .
Explanation
Problem Analysis We are given the quadratic equation 2 x 2 − 5 x + 1 = 3 . Our goal is to solve for x .
Rewriting 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 3 from both sides of the equation:
2 x 2 − 5 x + 1 − 3 = 0
This simplifies to:
2 x 2 − 5 x − 2 = 0
Applying the Quadratic Formula Now 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
In our equation, a = 2 , b = − 5 , and c = − 2 . Plugging these values into the quadratic formula, we get:
x = 2 ( 2 ) − ( − 5 ) ± ( − 5 ) 2 − 4 ( 2 ) ( − 2 )
Simplifying the Expression Now, we simplify the expression:
x = 4 5 ± 25 + 16
x = 4 5 ± 41
Final Answer Therefore, the solutions for x are:
x = 4 5 + 41 and x = 4 5 − 41
So, the final answer is:
x = 4 5 ± 4 41
Examples
Quadratic equations are incredibly useful in various real-world scenarios. For instance, they can model the trajectory of a ball thrown in the air, helping to determine its maximum height and range. Similarly, engineers use quadratic equations to design parabolic mirrors and reflectors, optimizing the focus of light or radio waves. Understanding how to solve these equations is fundamental in physics, engineering, and even economics, where they can model cost and revenue curves to find break-even points.
