Question

solve for y.
$9y^{2}-48y + 64 = 0$
write each solution as an integer, proper fraction, or improper fraction in simplest form.
if there are multiple solutions, separate them with commas.
$y = $

Explanation:

Step1: Identify the quadratic form

The equation is \(9y^2 - 48y + 64 = 0\), which is a quadratic in the form \(ay^2+by+c = 0\) with \(a = 9\), \(b=- 48\), \(c = 64\). We can try to factor it or use the quadratic formula. Let's try factoring. We need two numbers that multiply to \(a\times c=9\times64 = 576\) and add up to \(b=-48\). Notice that \(-24\times - 24=576\) and \(-24+-24=-48\). So we can rewrite the middle term:
\(9y^2-24y - 24y+64 = 0\)

Step2: Factor by grouping

Group the first two and last two terms:
\((9y^2 - 24y)+(-24y + 64)=0\)
Factor out the greatest common factor from each group:
\(3y(3y - 8)-8(3y - 8)=0\)
Now we can factor out \((3y - 8)\):
\((3y - 8)(3y - 8)=0\) or \((3y - 8)^2=0\)

Step3: Solve for y

Set \(3y - 8=0\)
Add 8 to both sides: \(3y=8\)
Divide both sides by 3: \(y=\frac{8}{3}\) (since the square means the root is repeated)

Answer:

\(\frac{8}{3}\)