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!

 May 5, 2023

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!

 May 5, 2023

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?

 May 5, 2023

Don't get into an argument with an idiot.

They drag you down to their level, then beat you on experience.

Guest May 5, 2023

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