dat practice math question

[tlcmqn]

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The sum of the squares of four consecutive integers is 366. Approximately what is the average of the three greatest squares?

how would you set this up?

 

[USCbiograd]

I think it goes something like this:

(x^1) + (x+1)^2 + (x+2)^2 + (x+3)^2 = 366

Set up to use in Quadratic formula.

This could definetly be wrong though, my math skills are a little rusty. If you have the solution, sometimes it can be easier to work back from that.

Edit: I previously Calclated the wrong number, but you should be able to get the correct answer this way

 

[Streetwolf]

12x, not 6x.

Also it wants the average of the SQUARES.

Would solve more but gotta go to class.

 

[klutzy1987]

I would do it a little differently. I would set it up as (x^2) + (x+1)^2 + (x+2)^2 + (X+3)^2 = 366. Solve for x which would yield 4X^2 + 12X - 352=0 Dividing by 4 all ways gives x^2+3x-88=0. Solve for X gives x=11 or x=-8. So there would give 2 solutions for this problem, either if you use 11 it would be (144+169+196)/3=169.66. If you use -8 it would be (49+36+25)/3=36.66.

 

[Streetwolf]

The sum of the squares of four consecutive integers is 366. Approximately what is the average of the three greatest squares?

how would you set this up?

x^2 + (x+1)^2 + (x+2)^2 + (x+3)^2 = 366

4x^2 + 12x + 14 = 366
4x^2 + 12x - 352 = 0
4(x^2 + 3x - 88) = 0
4(x-8)(x+11) = 0
x = 8, -11

(Not -8 and 11 as Klutzy wrote - you swapped them)

Notice how this gives 8,9,10,11 or -11,-10,-9,-8. Thus when you square them and take the largest 3 squares it will be 9^2, 10^2, and 11^2.

This is 81 + 100 + 121 = 302.

Take 302 / 3 and get 100.67.

 

[Marmar tala]

It's easier to have:
(n-1)^2 +n^2 + (n+1)^2 +(n+2)^2 = 366
4n^2 + 4n +6 = 366
2n^2+2n +3 = 183 => 2n^2 + 2n = 180 => n^2 +2n = 90 => n(n+1)=90
=> n=9
the average of the three:
(n^2 + (n+1)^2 +(n+2)^2 )/3 = (3 n^2 + 6n + 5)/3 = 99+5/3 ~100.67

 

[gabe85]

Okay...I suck at math and here this goes, approximate. 10 squared is 100. So the 4 consecutive numbers have to be around 10ish. If you include too many numbers past 10 its greater than 366. Guess and check. 8 squared = 64, 9 squared= 81, 10 squared = 100, 11 squared = 121. Add these up = 366. Then take the average of the last 3.

No need for equation IMO.😎

 

[jdm789]

Okay...I suck at math and here this goes, approximate. 10 squared is 100. So the 4 consecutive numbers have to be around 10ish. If you include too many numbers past 10 its greater than 366. Guess and check. 8 squared = 64, 9 squared= 81, 10 squared = 100, 11 squared = 121. Add these up = 366. Then take the average of the last 3.

No need for equation IMO.😎

thats how i did this problem. i initially thought it was 6,7,8,9 but it was way too low, so i decided to do 8,9,10,11 and it was right. using a formula is too time consuming, just guess and check.

 

[klutzy1987]

(Not -8 and 11 as Klutzy wrote - you swapped them)

Ugh thats why you dont try solving quadratics in your head.