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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 $(4,-9)$. The quadratic $ax^2 + bx +c$  has two real roots. The greater root is $\sqrt{n}+3$. What is $n$?

 Aug 24, 2023
 #1
avatar+128732 
+1

We have these two equations

 

1 = a (1 - 2)^2 + k   →    1 = a + k        (1)

-9 = a (4 - 2)^2 + k  →    -9 = 4a +k      (2)

 

Subtract (2) from (1)

 

10 = -3a

a = -10/3

 

k = 1 - a  = 1 - (-10/3) = 13/3

 

So the polynomial is

 

y = (-10/3)(x - 2)^2  + 13/3

 

And we  are trying to  find the  roots

 

So

 

(-10/3) ( x - 2)^2  + 13/3  = 0                     multiply through by 3

 

-10(x - 2)^2 + 13 =  0

 

(x - 2)^2  =  13/10            take the  positive root

 

x - 2 = sqrt (13/10)

 

x =  2 + sqrt (13/10)

 

Check your question.....your  root is not  correct

 

cool cool cool

 Aug 24, 2023
 #2
avatar+177 
+1

If, for some reason, you are convinced that the question is worded correctly despite the evidence otherwise, you can still find a value for n that will satisfy this problem. I will pick up where Cphill left off.

 

\( x = 2 + \sqrt{\frac{13}{10}} \\ 2 + \sqrt{\frac{13}{10}} = \sqrt{n} + 3 \\ \sqrt{n} = -1 + \sqrt{\frac{13}{10}} \\ n = 1 - 2\sqrt{\frac{13}{10}} + \frac{13}{10} \\ n = \frac{23}{10} - 2\sqrt{\frac{13}{10}}\)

 

However, this is quite an odd answer given the question.

 Aug 25, 2023

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