Answer:
[tex] \sqrt{32} + 3[/tex]
[tex] - \sqrt{32} + 3[/tex]
Step-by-step explanation:
Use foil method
[tex](y - 3) {}^{2} [/tex]
[tex](y - 3)( y - 3)[/tex]
First multiply y and y
[tex]y \times y = {y}^{2} [/tex]
Next Multiply Outer term y and -3.
[tex]y \times - 3 = - 3y[/tex]
After that ), Multily inner tern -3 and y
[tex] - 3 \times y = - 3y[/tex]
Finally multiply -3 and -3 and you get 9.
Then combine all
[tex] {y}^{2} - 6y + 9 - 32 [/tex]
[tex] {y}^{2} - 6y - 23[/tex]
Now I'm going to complete the square. First get only the y value on the left side so we first set this equal to zero then move add 23 on both sides.
[tex] {y}^{2} - 6y - 23 = 0[/tex]
Add 23
[tex]y {}^{2} - 6y = 23[/tex]
Apply formula
[tex] (\frac{b}{2} ) {}^{2} [/tex]
where b is the coefficient.
plug it in
[tex]( \frac{ - 6}{2}) {}^{2} [/tex]
Which is -3 squared which equal 9.
Then add 9 on both sides
[tex] {y}^{2} - 6y + 9 = 23 + 9[/tex]
23+9=32
Now let factor the left side
the only number that adds to -6 and multiply to 9 is
-3 so we would rewrite the left side as
[tex](x - 3) {}^{2} = 32[/tex]
Now take the sqr root of 32
[tex] \sqrt{(x - 3) {}^{2} } = \sqrt{32} [/tex]
Then add 3 on both sides
[tex]x + - 3 + 3 = \sqrt{32} + 3[/tex]
A sqr root of a number can be positive or negative so
the answer is
[tex] - \sqrt{32} + 3[/tex]
[tex] \sqrt{32} + 3[/tex]
