SOLUTION: Hi, In my Algebra II class we are working on Solving Quadratic Equations by graphing. I relize how to work the problems with a plain vertex like in this problem x2+2x-8=0 the verte
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-> SOLUTION: Hi, In my Algebra II class we are working on Solving Quadratic Equations by graphing. I relize how to work the problems with a plain vertex like in this problem x2+2x-8=0 the verte
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Question 28745: Hi, In my Algebra II class we are working on Solving Quadratic Equations by graphing. I relize how to work the problems with a plain vertex like in this problem x2+2x-8=0 the vertex is 1 and so the root/answer would be (-4 and 2) but i am unsure of solving the ones with a fraction for the vertex like in this problem x2-5x+4=0 and the vertex is 5/2 could you please help me?
thanks so much,
Chelsea Answer by venugopalramana(3286) (Show Source):
You can put this solution on YOUR website! Hi, In my Algebra II class we are working on Solving Quadratic Equations by graphing. I relize how to work the problems with a plain vertex like in this problem x2+2x-8=0
(X+1)^2-9=0 = (X+1)^2- (3)^2
the vertex is 1 ....NO....PLEASE CORRECT
THE VERTEX IS AT X=-1 AND Y = -9....
and so the root/answer would be (-4 and 2)...THAT IS -1+3=2 AND -1-3=-4
OK...GOOD..SEE THE GRAPH BELOW...
but i am unsure of solving the ones with a fraction for the vertex like in this problem x2-5x+4=0
(X-5/2)^2 - 9/4 = 0 = (X-5/2)^2 - (3/2)^2
and the vertex is 5/2
CORRECT ..YOU ARE CORRECT HERE..HOW DID YOU MISTAKE IN THE EARLIER CASE?
could you please help me?
SO THE ROOTS ARE 5/2 + 3/2 =8/2 = 4.......OR.....5/2 - 3/2 = 2/2 =1...THAT IS 4 OR 1.
I HOPE YOU GOT THE METHOD..THE ROOTS ARE OBTAINED BY ADDING /SUBTRACTING THE SQUARE ROOT OF CONSTANT TERM FROM THE VERTEX (OR PRECISELY X COORDINATE OF VERTEX)
SEE THE GRAPH BELOW
