Answer :
Solving y²-14y=0 by splitting the middle
We want to make sure that we have no y² coefficient and put any constants on the right side. Check.
Now, take half of the y coefficient, (half of -14 would be -7) square it, (-7² = 49) and add it to both sides.
[tex]y^2-14y+49=49[/tex]
Factor the left side. (Since the method we're using sets up a perfect square trinomial, it's going to be (y+half of the y coefficient)².
[tex](y-7)^2=49[/tex]
Take the square root of each side.
[tex]y-7=\±7[/tex]
[tex]y=7\±7[/tex]
[tex]\boxed{y=0\ or\ 14}[/tex]
We want to make sure that we have no y² coefficient and put any constants on the right side. Check.
Now, take half of the y coefficient, (half of -14 would be -7) square it, (-7² = 49) and add it to both sides.
[tex]y^2-14y+49=49[/tex]
Factor the left side. (Since the method we're using sets up a perfect square trinomial, it's going to be (y+half of the y coefficient)².
[tex](y-7)^2=49[/tex]
Take the square root of each side.
[tex]y-7=\±7[/tex]
[tex]y=7\±7[/tex]
[tex]\boxed{y=0\ or\ 14}[/tex]
y² - 14y = 0
y² - 14y - 0 = 0
x = -(-14) +/- √(-14)² - 4(1)(0))
2(1)
x = 14 +/- √(196 - 0)
2
x = 14 +/- √(196)
2
x = 14 +/- 14
2
x = 7 + 7
x = 7 + 7 x = 7 - 7
x = 14 x = 0
y² - 14y - 0 = 0
x = -(-14) +/- √(-14)² - 4(1)(0))
2(1)
x = 14 +/- √(196 - 0)
2
x = 14 +/- √(196)
2
x = 14 +/- 14
2
x = 7 + 7
x = 7 + 7 x = 7 - 7
x = 14 x = 0
