Question

what are the values of x in the equation $x^2 - 6x + 9 = 25$?
$\circ$ $x = -2$ or $x = 8$
$\circ$ $x = -1$ or $x = -11$
$\circ$ $x = 1$ or $x = 11$
$\circ$ $x = 2$ or $x = -8$

Explanation:

Step1: Rearrange the equation

First, we start with the equation \(x^{2}-6x + 9=25\). Subtract 25 from both sides to get it in standard quadratic form: \(x^{2}-6x + 9 - 25=0\), which simplifies to \(x^{2}-6x - 16 = 0\). Alternatively, we can recognize that the left - hand side \(x^{2}-6x + 9\) is a perfect square trinomial, since \((a - b)^2=a^{2}-2ab + b^{2}\), and here \(a = x\), \(b = 3\), so \(x^{2}-6x + 9=(x - 3)^{2}\). So our equation becomes \((x - 3)^{2}=25\).

Step2: Solve for x

Take the square root of both sides. Remember that if \(y^{2}=a\) (\(a\geq0\)), then \(y=\pm\sqrt{a}\). So for \((x - 3)^{2}=25\), we have \(x - 3=\pm\sqrt{25}=\pm5\).

Case 1: When \(x - 3 = 5\), add 3 to both sides: \(x=5 + 3=8\).

Case 2: When \(x - 3=-5\), add 3 to both sides: \(x=-5 + 3=-2\).

Answer:

\(x=-2\) or \(x = 8\) (corresponding to the first option: \(x=-2\) or \(x = 8\))