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 #1
avatar+118696 
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Feb 21, 2018
 #8
avatar+118696 
+2

I am quite sure EP will be correct but I will take a look:

 

Let b and c be constants such that the quadratic 2x2+bx+c   

has roots 3+5  and 35.

Find the vertex of the graph of the equation   

y=2x2+bx+c

 

The roots of a quadratic   ax2+bx+c  is the answers to   ax2+bx+c=0

And the answers to this are given by the quadratic equation

 

x=b±b24ac2ax=b2a±b24ac2a=3±5

 

This means that  x=b2a = 3      must be exactly halfway between the two roots.!!

 

So the axis of symmetry is x=3 and the vertex lies on this line so the x value of the vertex is 3

The y value will be    y=232+bx+c=18+3b+c

 

This is fine but you need to find the vleu of b and c       frown

 

From the equation above I can see that 

b2a=3and±b24ac2a=±5a=2sob4=3and±b2+8c4=±5b=12and±b2+8c4=±5b=12andb2+8c16=5b=12andb2+8c=80b=12and144+8c=80b=12and18+c=10b=12andc=8

 

so

 

y=18+3b+cy=18+312+8y=18+368y=10

 

So the vertex is   (3,10)

 

Which is exactly the same answer has given you. He has done it a number of different ways but stlill all the answers are the same!!

 

Thanks EP :)

Feb 21, 2018

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