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A parabola with equation $y=ax^2+bx+c$ has a vertical line of symmetry at $x=2$ and goes through the two points $(1,1)$ and $(4,-7)$. The quadratic $ax^2 + bx +c$ has two real roots. The greater root is $\sqrt{n}+2$. What is $n$? So far I have tried replacing x and y For instance, for the point (1,1), I tried 1=a+b+c while for (4,-7) I tried 4=49a-7b+c

 Aug 19, 2022
 #2
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If the vertical line of symmetry   is x =2

Then  the x coordinate of the  vertex  =  2  =    -b /(2a)

So    -b / (2a)  = 2       

So  - b  = 4a

So  b =  -4a

 

And we have this system

 

a (1)^2 + (-4a)(1) + c  = 1

a(4)^2 + (-4a)(4) + c  =  -7                      simplify this system

 

a - 4a + c  = 1   ⇒   -3a + c  =1

16a - 16a  + c =  -7  ⇒  c  = -7

 

c =  -7

 

So

 

-3a  -  7  =  1

-3a  = 8

a = -8/3

 

b =  -4 (-8/3)  = 32/3

 

The quadratic is    y =    -(8/3)x^2  + (32/3)x - 7

 

So

 

-(8/3) x^2 + (32/3)x  - 7  =   0          mulriply through by 3

 

-8x^2  + 32x - 21  =   0

 

The  greater root   = 

 

- 32  -  sqrt ( 32^2  - 4(-21)(-8))

_________________________

           2 (-8)

 

-32  - sqrt (352)

_____________

          -16

 

-32 -  4sqrt(22)

____________

          -16

 

2  + (1/4)sqrt (22)   =    2 + sqrt (22/16)  =  2 + sqrt (11/8)

 

n =   11/8

 

Here's a graph  :  https://www.desmos.com/calculator/feiavcp455

 

 

cool cool cool

 Aug 19, 2022

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