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Given that x^2 +ax +b =(x+2)^2 -9 work out a and b

 

expression x^2 - 10x - 5 can be written in form (x+p)^2 + q

find value of p and q

 

write expression 2x^2 - 8x + 19 in form a(x+b)^2 + c

 

find coordinates of minimum point on graph

 

state if and where the graph crosses x-axis

Guest Oct 25, 2017
 #1
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Given that x^2 +ax +b =(x+2)^2 -9 work out a and b

 

Just expand (x+2)^2 -9  = 

 

x^2 + 4x + 4 - 9  =  x^2 + 4x - 5    ⇒ a = 4      b = -5

 

 

 x^2 - 10x - 5

 

Take  1/2 of 10  = 5.....square it  = 25 .... add and subtract it

 

x^2 - 10x + 25 -  25           factor the first three terms

 

(x - 5)^2  - 25     =  ( x + (-5) )^2   -  25    ⇒    p =  -5        q  = -25   

 

 

 

write expression 2x^2 - 8x + 19 in form a(x+b)^2 + c

 

Factor out  2

 

2 [  x^2  - 4x  +  19/2]

 

Take 1/2 of 4  = 2......square it......= 4.....add and subtract it

 

2 [  x ^2 -  4x  + 4   +  19/2  -  4 ]

 

2 [  x^2 - 4x  +  4  +  19/2  - 8/2 ]      factor the first three terms

 

2 [ ( x - 2)^2  +  11/2] 

 

2(x - 2)^2  +  11

 

The minimum point on the graph  is   (2,11)

 

Since this minimum lies above the x axis......the graph never crosses that axis

 

 

 

cool cool cool

CPhill  Oct 25, 2017

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