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
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))
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2 (-8)
-32 - sqrt (352)
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-16
-32 - 4sqrt(22)
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-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