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?

HumenBeing Jul 19, 2023

#1**0 **

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

Therefore, the answer is 25/4.

Guest Jul 19, 2023

#3**0 **

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.

HumenBeing Jul 19, 2023

#4**0 **

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.

HumenBeing Jul 19, 2023

#6**0 **

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 Jul 19, 2023

#8**-2 **

"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**+1 **

I still wonder when this website turned from a math website to an insulting website.

history
Jul 20, 2023

#12**0 **

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

#7**0 **

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.

Tiggsy Jul 20, 2023

#14**0 **

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 (***):

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

Guest Jul 23, 2023