What value(s) of $x$ make the equation $x^2-18x+81=0$ true?
The given equation is \(x^2 - 18x + 81 = 0\). This is a quadratic equation of the form \(ax^2 + bx + c = 0\), where \(a=1\), \(b=-18\), and \(c=81\). To solve this, we can look for ways to factor the expression \(x^2 - 18x + 81\).
We observe that the constant term \(81\) is a perfect square, \(9^2\). The middle term \(-18x\) is equal to \(-2 \times 9 \times x\). This structure matches the pattern of a perfect square trinomial, \(a^2 - 2ab + b^2 = (a - b)^2\). Specifically, with \(a=x\) and \(b=9\), we have \(x^2 - 2(x)(9) + 9^2 = (x-9)^2\). Therefore, the equation \(x^2 - 18x + 81 = 0\) can be rewritten as \((x-9)^2 = 0\).
To find the value of \(x\), we take the square root of both sides of the equation \((x-9)^2 = 0\). This gives \(x-9 = 0\). Adding 9 to both sides, we find \(x = 9\). Comparing this result to the given options, we find that it corresponds to option b.
Quadratic equations are widely used in physics and engineering to model movement. For example, if you throw a ball, its height over time is often described by a quadratic equation. Solving these equations helps you find when the ball hits the ground or reaches its peak height. By identifying perfect square patterns, you can simplify complex-looking problems and solve them efficiently, just as we did here to find the unique solution.