Algebraic Booster #1

$(a^{2}-a)+ (b^{2}-b) + (c^{2}-c) + (d^{2}-d)+ 1 = 0$

$(a^{2}-a+\frac{1}{4})+ (b^{2}-b+\frac{1}{4}) + (c^{2}-c+\frac{1}{4}) + (d^{2}-d+\frac{1}{4}) = 0$

$(a-\frac{1}{2})^{2} + (b-\frac{1}{2})^{2} + (c-\frac{1}{2})^{2} + (d-\frac{1}{2})^{2} = 0$

The value inside the square will always be zero or positive, so if all 4 expressions inside the brackets add up to 0, the value of each expression inside the bracket must be zero.

So, $a = \frac{1}{2}, b= \frac{1}{2}, c = \frac{1}{2}, d = \frac{1}{2}$

So,

$a + b + c + d = 2$

I tried this one but I dont know if it would be reasonable.

Let $a = b = c = d = x$ then $a + b + c + d = 4x$ and $a^2 + b^2 + c^2 + d^2 = 4x^2$ .

As a result, we are looking for the value of $4x$ .

Now we have:

$4x^2 - 1 = 4x$ or $4x^2 -4x +1 = 0$ which can be factored into $(2x -1)^2 = 0$ This gives us $x = \frac{1}{2}$ . Thus, $4x = a + b + c + d = 4 (\frac{1}{2}) = 2$