+214 The real numbers a, b, c, and d satisfy \(a^2 + b^2 + c^2 + 1 = d + \sqrt{a + b + c - d}\).
Find the value of d.
I know that there is some sort of quadratic or factorization identity that will make this problem easy, but I do not know what it is.
Thanks in advance to potential answerers!
The equation a^2+b^2+c^2+1=d+sqrt(a+b+c−d) can be rewritten as a^2+b^2+c^2+1−d=sqrt(a+b+c−d). Squaring both sides, we get a^2+b^2+c^2+1−2d+d^2=a+b+c−d. Simplifying, we get a^2+b^2+c^2+d^2−2d=a+b+c.
Now, we can use the fact that a^2+b^2+c^2=1 to substitute for a^2+b^2+c^2 in the equation above. This gives us 1+d^2−2d=a+b+c.
Finally, we can use the fact that a+b+c−d=0 to substitute for a+b+c in the equation above. This gives us 1+d^2−2d=0.
Solving for d, we find that d = (1+/- sqrt(5))/2.
However, we know that d must be a real number, so the only possible value of d is (1+sqrt(5))/2.
Therefore, the value of d is (1+sqrt(5))/2, which is the golden ratio. Nice!
+214 I have a few questions on that answer.
When I square \(a^2+b^2+c^2+1-d\), I get \(a^4+b^4+c^4+a^2b^2+a^2c^2+2a^2-2a^2d+b^2a^2+b^2c^2+2b^2-2b^2d+c^2a^2+c^2b^2+2c^2-2c^2d+d^2-2d+1\).
How did you get \(a^2+b^2+c^2+1−2d+d^2\)?
Where did you get \(a^2+b^2+c^2=1\)?
Where did you get \(a+b+c-d=0\)?
And if those are both true, then wouldn't d be 2?
