Question

consider the quadratic equation: $x^2 - 10x = -25$
a. using the method for completing the square, transform the quadratic equation into the square of a binomial on one side of the equation and a constant on the other side.
your answer should look like: $(x - a)^2 = b$.
the equation is:
b. using your equation in (a), find all solutions to the quadratic equation. write answers in exact form.
separate multiple answers with a comma and write complex solutions in terms of $i$.
$x = $

Explanation:

Part a

Step1: Rewrite the equation

The given equation is \(x^{2}-10x=-25\). To complete the square for the quadratic expression \(x^{2}-10x\), we take half of the coefficient of \(x\), square it, and add it to both sides. The coefficient of \(x\) is \(- 10\), half of it is \(\frac{-10}{2}=-5\), and squaring it gives \((-5)^{2} = 25\).

Step2: Complete the square

Add \(25\) to both sides of the equation \(x^{2}-10x=-25\):
\(x^{2}-10x + 25=-25 + 25\)
The left - hand side is a perfect square trinomial, and we know that \(x^{2}-10x + 25=(x - 5)^{2}\), and the right - hand side is \(0\). So the equation becomes \((x - 5)^{2}=0\).

Step1: Take the square root of both sides

We have the equation \((x - 5)^{2}=0\). Taking the square root of both sides, we get \(\sqrt{(x - 5)^{2}}=\sqrt{0}\).

Step2: Solve for \(x\)

Since \(\sqrt{(x - 5)^{2}}=\vert x - 5\vert\) and \(\sqrt{0} = 0\), we have \(\vert x - 5\vert=0\). The absolute value equation \(\vert x - 5\vert=0\) implies that \(x-5 = 0\). Solving for \(x\), we add \(5\) to both sides of the equation: \(x=5\). Since the square root of \(0\) has only one solution (a repeated root), the solution is \(x = 5\) (with multiplicity \(2\)).

Answer:

\((x - 5)^{2}=0\)

Part b