Question

solve for y.
$y^{2}-18y + 81 = 0$
write each solution as an integer, proper fraction, or improper fraction in simplest form. if there are multiple solutions, separate them with commas.
$y = \square$
submit

Explanation:

Step1: Recognize the quadratic form

The equation \(y^2 - 18y + 81 = 0\) is a quadratic equation. Notice that it can be factored as a perfect square trinomial. Recall that \((a - b)^2 = a^2 - 2ab + b^2\). Here, \(a = y\), and \(2ab = 18y\), so \(2b = 18\) which means \(b = 9\), and \(b^2=81\). So we can factor the left - hand side as \((y - 9)^2=0\).

Step2: Solve for y

If \((y - 9)^2 = 0\), then by the zero - product property (if \(A^2 = 0\), then \(A = 0\)), we set \(y-9 = 0\). Solving for \(y\), we add 9 to both sides of the equation: \(y=9\).

Answer:

9