Solve the equation [tex]$x ^2- 4 x + 6 = 0$[/tex] by completing the square
Answer (2)
We solved the equation x 2 − 4 x + 6 = 0 by completing the square. The solutions are x = 2 + i 2 and x = 2 − i 2 . This method involves rearranging the equation, completing the square, and utilizing imaginary numbers due to the negative square root.
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Rewrite the equation: x 2 − 4 x = − 6 .
Complete the square: x 2 − 4 x + 4 = − 6 + 4 , which simplifies to ( x − 2 ) 2 = − 2 .
Take the square root: x − 2 = ± − 2 = ± i 2 .
Solve for x : x = 2 ± i 2 .
x = 2 ± i 2
Explanation
Understanding the Problem We are given the quadratic equation x 2 − 4 x + 6 = 0 , and our goal is to solve it by completing the square. This method involves transforming the equation into a form that allows us to easily find the solutions for x .
Isolating the x Terms First, we rewrite the equation by moving the constant term to the right side: x 2 − 4 x = − 6
Completing the Square To complete the square, we need to add a value to both sides of the equation that will make the left side a perfect square trinomial. We take half of the coefficient of the x term, which is − 4 , and square it: ( 2 − 4 ) 2 = ( − 2 ) 2 = 4 . So, we add 4 to both sides of the equation: x 2 − 4 x + 4 = − 6 + 4
Rewriting as a Square Now, we can rewrite the left side as a squared term: ( x − 2 ) 2 = − 2
Taking the Square Root Next, we take the square root of both sides of the equation: x − 2 = ± − 2
Introducing Imaginary Unit Since we have a negative number under the square root, we introduce the imaginary unit i , where i = − 1 . Thus, − 2 = 2 ⋅ − 1 = i 2 . So we have: x − 2 = ± i 2
Solving for x Finally, we solve for x by adding 2 to both sides: x = 2 ± i 2
Final Answer Therefore, the solutions to the equation x 2 − 4 x + 6 = 0 are x = 2 + i 2 and x = 2 − i 2 .
Examples
Completing the square is a useful technique in physics, especially when dealing with simple harmonic motion or analyzing the motion of projectiles. For example, when analyzing the height of a projectile over time, the equation often takes a quadratic form. By completing the square, we can easily determine the maximum height reached by the projectile and the time at which it occurs. This method transforms the quadratic equation into vertex form, providing direct insights into the key characteristics of the projectile's trajectory, such as its peak and symmetry.
