Algebra Question involving variables?

You have three equations in four unknowns, so there may well be an infinite

number of solutions. At least two exist, as demonstrated by francis and Hemant.

The description is of four equally-spaced numbers where either the largest or the

smallest is the sum of squares of the the other three. If δ is the distance

between consecutive numbers, this sounds like there may be a solution for many

values of δ = b-a = c-b = d-a; namely:

d = a² + b² + c²

d = (d-3δ)² + (d-2δ)² + (d-δ)²

d = (d² - 6δd + 9δ²) + (d² - 4δd + 4δ²) + (d² - 2δd + δ²)

d = 3d² - 12δd + 14δ²

3d² - (12δ + 1)d + 14δ² = 0

That has a solution whenever:

(12δ + 1)² - 4(3)(14δ²) >= 0

(144δ² + 24δ + 1) - 168δ² >= 0

-24δ² + 24δ² + 1 >= 0

δ² - δ - 1/24 <= 0

The roots are at

δ = (1/2)[1 +/- sqrt(1 + 4/24)]

δ = (1/2)[1 +/- sqrt(7/6)] = 1/2 +/- sqrt(42)/12

Since 6 < sqrt(42) < 7, the only integer values of δ with solutions are 0 and 1,

which include the first two solutions mentioned. The other solutions with these delta values are

δ = 0: 3d² - d = 0 ==> d= 0 (given by francis) or d=1/3

(1/3)² + (1/3)² + (1/3)² = 3/9 = 1/3

...so a=b=c=1/3 is the other solution with all numbers equal.

δ = 1: 3d² - 13d + 14 = 0 = (3d - 7)(d - 2)

So d=2 is the solution mentioned by Hemant, and d=7/3 is another:

(-2/3)² + (1/3)² + (4/3)² = 4/9 + 1/9 + 16/9 = 21/9 = 7/3

...is the other solution with unit spacing between numbers.

As nudnik0 demonstrated while I was typing and debugging this, (or you can see from the equation above too) the solutions for (d,δ) lie on an ellipse. The solutions given by Hemant and francis correspond to the only integer points on that ellipse.

If we let x := d - c = c - b = b - a be the common difference and m := (a + b + c + d) / 4 be the mean value of a, b, c, and d, then the only remaining constraint is the equation a^2 + b^2 + c^2 = d, which reduces to

(m - 3x / 2)^2 + (m - x / 2)^2 + (m + x / 2)^2 = m + 3x / 2

<==> 3 m^2 - 3 x m + 11 x^2 / 4 = m + 3x / 2.

This is an ellipse in the (m, x)-plane and does not give enough information to uniquely find m. Therefore, the equations given do not determine a + b + c + d. For example, one solution is

a = -0.6117957, b = 0.3684219, c = 1.3486394, d = 2.3288570

==> a + b + c + d = 3.4341226

and another is

a = -0.0548709, b = -0.0350885, c = -0.0153061, d = 0.0044763

==> a + b + c + d = -0.1007893.

Given 4 numbers a, b, c, d such that

d - c = c - b = b - a and

a² + b² + c² = d

Find the value of a + b + c + d

d - c = c - b

2c = b + d

c - b = b - a

2b = a + c

d - c = b - a

a + d = b + c

a + b + c + d = 2(b + c)

a + b + c + a² + b² + c² = 2(b + c)

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

(a + b + c)²

= a² + b² + c² + 2ab + 2ac + 2bc

hi buddy this an person-friendly question in line with linear equations you ought to use transposition technique in this equipment whilst a term is transfered from one ingredient to different its sign turns into opposite answer: x+4-7x=22 =4-7x=22-x (right here x comes from the left ingredient its sign variations) =4=22-x+7x(right here 7x comes from the left ingredient its sign variations) =4-22=6x(comparable) =-18=6x x=-18/6 =-3 Bye Bye!!!!!!!!!!!!!!!!!!!!!!

hi buddy this an elementary question consistent with linear equations you need to use transposition approach for the time of this methodology while a term is transfered from one area to different its sign turns into opposite answer: x+4-7x=22 =4-7x=22-x (here x comes from the left area its sign differences) =4=22-x+7x(here 7x comes from the left area its sign differences) =4-22=6x(same) =-18=6x x=-18/6 =-3 Bye Bye!!!!!!!!!!!!!!!!!!!!!!

a = -1, b = 0, c = 1, d = 2

a + b + c + d = -1 + 0 + 1 + 2 = 2 .......... Ans.

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0, they're all 0's LOL