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 $n$?

Guest Aug 24, 2023

#1**+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

CPhill Aug 24, 2023

#2**+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.

The3Mathketeers Aug 25, 2023