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 $n$?
+129895 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
+189 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.
