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# I challenge someone to solve this (correctly)

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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 √n + 2. What is n?

Jul 19, 2023

#1
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The vertical line of symmetry means the parabola is symmetric around the line x = 2, so the x-coordinate of the vertex is 2. Since the vertex is the midpoint of the segment connecting the two points (1,1) and (4,-7), the x-coordinate of the vertex is 21+4​=3. Therefore, the vertex of the parabola is (3,v), where v is the value of y when x = 3.

We know that the parabola goes through the point (1,1), so we can use the point-slope form of the equation of a line to get the equation of the parabola:

y - 1 = m(x - 1)

where m is the slope of the parabola. Since the parabola is symmetric around the line x = 2, the slope of the parabola is equal to the negative reciprocal of the slope of the line connecting the two points (1,1) and (4,-7), which is (-7 - 1)/(4 - 1) = -8/3. Substituting this into the point-slope form of the equation of the parabola, we get the equation of the parabola:

y - 1 = (-8/3)(x - 1)

or

y = -8/3 * x + 17/3

To find the value of v, we substitute x = 3 into this equation to get

v = -8/3 * 3 + 17/3 = 1

Therefore, the vertex of the parabola is (3,1).

Since the greater root of the quadratic is √n + 2, the other root must be √n - 2. Since the sum of the roots of the quadratic is 3 (the x-coordinate of the vertex), we have

√n - 2 + √n + 2 = 3

or

2√n = 5

or

√n = 5/2

or

n = (5/2)^2 = 25/4

Jul 19, 2023
#2
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This is chatgpt generated, but it won't get deleted since "we don't have any proof" or some other st*pid reason. But I don't care, in fact I encourage the use of ChatGPT to confuse homework cheaters such as HumenBeing

Guest Jul 19, 2023
#3
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This isn't my hw though... and I could tell it was ChatGPT generated just by looking at it. Not every question asked is from homework.

Jul 19, 2023
edited by HumenBeing  Jul 19, 2023
edited by HumenBeing  Jul 19, 2023
#4
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And I noticed the amount of disinformation here, there are a lot of good answers, but also a lot of ChatGPT generated and wrong answers... That's not what this website is for.

Jul 19, 2023
#5
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I am perfectly fine with disinformation for homework cheaters, the website is not for them.

Guest Jul 19, 2023
#6
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I don't cheat on my homework, it's way too easy I just solve it myself, it's not even worth the time to post a question.

Jul 19, 2023
#8
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"I don't cheat on my homework, it's way too easy I just solve it myself, it's not even worth the time to post a question. "

HumenBeing is trying to cover up his embarrassment of being a complete failure.

Guest Jul 20, 2023
#10
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I still wonder when this website turned from a math website to an insulting website.

history  Jul 20, 2023
#12
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His or her??? Don't assume gender, also are you sure I'm the failure here? If you care so much, why don't you at least get a registered account, more people would probably listen.

HumenBeing  Jul 22, 2023
#13
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Guess you're not talking to me, cause I'm female.

HumenBeing  Jul 22, 2023
#7
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The line of symmetry passes through the minimum point on the curve and that will be at x = -b/2a, so -b/2a = 2,

4a + b = 0 .....................(eq 1).

The curve passes through (1, 1), so

a + b + c = 1 .........................(eq 2)

and through (4 - 7) so

16a + 4b + c = -7 .....................(eq 3).

Solve the three equations simultaneously, begin by subtracting 4 times eq 1 from eq 3, to arrive at

a = -8/3, b = 32/3, c = -7.

Putting y = 0 and solving the resulting quadratic,

$$\displaystyle x = \frac{-32/3 \pm \sqrt{(32/3)^{2}-4(-8/3)(-7)}}{2(-8/3)} \\ \displaystyle =2 \mp(3/16)\sqrt{352/9} = 2 \mp(3/16)(4/3)\sqrt{22} \\ \displaystyle 2 \mp\sqrt{22/16}= 2 \mp \sqrt{11/8},$$

so n = 11/8.

Jul 20, 2023
#14
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Hi HumanBeing,

I am just confirming Tiggsy's amazing answer (must more efficient than my answer), and that the Guest's comment below his post is incorrect, in fact, n=11/8.

$$\text{Given } \space \space y=ax^2+bx+c \space \space \space \space \space (*) \\$$

$$\text{If the parabola passes through (1,1) and (4,-7)} \\ \text{Then, they must satisfy (*), thus substituting into the equation yields: } \\ \space\space\space\space\space\space\space\space\space1=a+b+c \space\space\space\space\space\space\space\space\space\space (1) \\ \space\space\space\space-7=16a+4b+c \space\space\space\space(2) \\$$

$$\text{Now, we have a system of two equations but with three unknowns} \\ \text{So, we have to use the last bit of information} \\ \text{Basically, if the greater root is } \space\space\space \sqrt{n} +2 = 2+\sqrt{n} \\ \text{Then, we know that irrational roots come in pairs, I.e. recall the following theorem: } \\ \text{if } \space\space a+\sqrt{b} \space\space\space \text{is a root, then} \space\space\space a-\sqrt{b} \space\space\space \text{must be a root of the polynomial}.$$

$$\text{Hence, } \space\space 2-\sqrt{n} \space\space \text{is a root} \\ \text{Sum of the roots: } \space \space 4 \\ \text{Product of the roots: } \space\space 4-n \\ \text{Apply Vieta's formulae: } -\dfrac{b}{a} = 4 \space\space\space (**) \\ \dfrac{c}{a}=4-n \space\space\space\space (***)\\$$

Using (**), we can eliminate b from (1) and (2) as follows:
$$b=-4a \\ \text{(1) and (2) becomes: } \space \\ \iff 1=a-4a+c \implies 1=-3a+c\\ \iff -7=16a-16a+c \implies c=-7 \\ \text{Thus, } a=-\dfrac{8}{3}$$

Using (***):

$$\dfrac{c}{a}=4-n \\ \iff \dfrac{-7}{\dfrac{-8}{3}}=4-n \\ \iff \dfrac{21}{8}=4-n \\ \iff n=1.375= \dfrac{11}{8} \\ \text{Which is the answer}.$$

Jul 23, 2023