Which equation is equivalent to the given equation?
$x^2-6 x=8$
The given equation is \(x^2 - 6x = 8\). Our goal is to transform this equation into the form \((x - h)^2 = k\), which is a process known as completing the square. Completing the square allows us to rewrite a quadratic expression of the form \(x^2 + bx\) as a perfect square trinomial plus or minus a constant. This is a powerful technique for solving quadratic equations and understanding their geometric properties.
To complete the square for the expression \(x^2 - 6x\), we need to add a constant to make it a perfect square trinomial. The general form of a squared binomial is \((x - h)^2 = x^2 - 2hx + h^2\). By comparing this with \(x^2 - 6x\), we identify that \(-2h = -6\), which means \(h = 3\). Thus, we need to add \(h^2 = 3^2 = 9\) to both sides of the original equation to maintain balance:
\[x^2 - 6x + 9 = 8 + 9\]
Adding 9 to both sides ensures the equality is preserved while creating a perfect square on the left.
Now, we can factor the left-hand side of the equation \(x^2 - 6x + 9\). Since \(x^2 - 6x + 9 = (x - 3)(x - 3) = (x - 3)^2\), we substitute this back into the equation. On the right-hand side, we simply calculate the sum \(8 + 9 = 17\). Putting these together, we get:
\[(x - 3)^2 = 17\]
This equation is now in the form \((x - h)^2 = k\). Comparing this to our original list of options, we find that this matches the first option.
Completing the square is not just an algebraic exercise; it is essential in geometry and physics. For example, when you have an equation for a circle like \(x^2 + y^2 - 6x + 8y = 0\), you must complete the square for both \(x\) and \(y\) to find the center and the radius of the circle. By converting it to \((x - 3)^2 + (y + 4)^2 = 25\), you immediately see the circle is centered at \((3, -4)\) with a radius of \(5\). This method helps in identifying paths, trajectories, and geometric shapes in various real-world applications.