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y^2-14y=0 solve the quadratic equation with steps please

Sagot :

RedRicinidn
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]
Panoyinidn
     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
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