APL (NARS), 2 bytes

√

NARS has built-in support for quaternions. ¯_(⍨)_/¯

Python 2, 72 bytes

def f(a,b,c,d):s=((a+(a*a+b*b+c*c+d*d)**.5)*2)**.5;print s/2,b/s,c/s,d/s

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More or less a raw formula. I thought I could use list comprehensions to loop over b,c,d, but this seems to be longer. Python is really hurt here by a lack of vector operations, in particular scaling and norm.

Python 3, 77 bytes

def f(a,*l):r=a+sum(x*x for x in[a,*l])**.5;return[x/(r*2)**.5for x in[r,*l]]

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Solving the quadratic directly was also shorter than using Python's complex-number square root to solve it like in the problem statement.

Wolfram Language (Mathematica), 19 bytes

Sqrt

<<Quaternions`

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Mathematica has Quaternion built-in too, but is more verbose.

_{Although built-ins look cool, do upvote solutions that don't use built-ins too! I don't want votes on questions reaching HNQ be skewed.}

JavaScript (ES7), 55 53 bytes

Based on the direct formula used by xnor.

Takes input as an array.

q=>q.map(v=>1/q?v/2/q:q=((v+Math.hypot(...q))/2)**.5)

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How?

Given an array $q=[a,b,c,d]$, this computes:

$$x=sqrt{frac{a+sqrt{a^2+b^2+c^2+d^2}}{2}}$$

And returns:

$$left[x,frac{b}{2x},frac{c}{2x},frac{d}{2x}right]$$

q =>                            // q = input array

  q.map(v =>                    // for each value v in q:

    1 / q ?                     //   if q is numeric (2nd to 4th iteration):

      v / 2 / q                 //     yield v / 2q

    :                           //   else (1st iteration, with v = a):

      q = (                     //     compute x (as defined above) and store it in q

        (v + Math.hypot(...q))  //     we use Math.hypot(...q) to compute:

        / 2                     //       (q[0]**2 + q[1]**2 + q[2]**2 + q[3]**2) ** 0.5

      ) ** .5                   //     yield x

  )                             // end of map()

Haskell, 51 bytes

f(a:l)|r<-a+sqrt(sum$(^2)<$>a:l)=(/sqrt(r*2))<$>r:l

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A direct formula. The main trick to express the real part of the output as r/sqrt(r*2) to parallel the imaginary part expression, which saves a few bytes over:

54 bytes

f(a:l)|s<-sqrt$2*(a+sqrt(sum$(^2)<$>a:l))=s/2:map(/s)l

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Java 8, 84 bytes

(a,b,c,d)->(a=Math.sqrt(2*(a+Math.sqrt(a*a+b*b+c*c+d*d))))/2+" "+b/a+" "+c/a+" "+d/a

Port of @xnor's Python 2 answer.

Try it online.

Explanation:

(a,b,c,d)->           // Method with four double parameters and String return-type

  (a=                 //  Change `a` to:

     Math.sqrt(       //   The square root of:

       2*             //    Two times:

         (a+          //     `a` plus,

          Math.sqrt(  //     the square-root of:

            a*a       //      `a`  squared,

            +b*b      //      `b` squared,

            +c*c      //      `c` squared,

            +d*d))))  //      And `d` squared summed together

  /2                  //  Then return this modified `a` divided by 2

  +" "+b/a            //  `b` divided by the modified `a`

  +" "+c/a            //  `c` divided by the modified `a`

  +" "+d/a            //  And `d` divided by the modified `a`, with space delimiters

Charcoal, 32 bytes

≔Ｘ⊗⁺§θ⁰ＸΣＥθ×ιι·⁵¦·⁵η≧∕ηθ§≔θ⁰⊘ηＩθ

Try it online! Link is to verbose version of code. Port of @xnor's Python answer. Explanation:

≔Ｘ⊗⁺§θ⁰ＸΣＥθ×ιι·⁵¦·⁵η

Square all of the elements of the input and take the sum, then take the square root. This calculates $ | x + yvec{u} | = sqrt{ x^2 + y^2 } = sqrt{ (a^2 - b^2)^2 + (2ab)^2 } = a^2 + b^2 $. Adding $ x $ gives $ 2a^2 $ which is then doubled and square rooted to give $ 2a $.

≧∕ηθ

Because $ y = 2ab $, calculate $ b $ by dividing by $ 2a $.

§≔θ⁰⊘η

Set the first element of the array (i.e. the real part) to half of $ 2a $.

Ｉθ

Cast the values to string and implicitly print.

05AB1E, 14 bytes

nOtsн+·t©/¦®;š

Port of @xnor's Python 2 answer.

Try it online or verify all test cases.

Explanation:

n                 # Square each number in the (implicit) input-list

 O                # Sum them

  t               # Take the square-root of that

   sн+            # Add the first item of the input-list

      ·           # Double it

       t          # Take the square-root of it

        ©         # Store it in the register (without popping)

         /        # Divide each value in the (implicit) input with it

          ¦       # Remove the first item

           ®;     # Push the value from the register again, and halve it

             š    # Prepend it to the list (and output implicitly)

Wolfram Language (Mathematica), 28 bytes

{s=#+Norm@{##},##2}/(2s)^.5&

Port of @xnor's Python 2 answer.

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C# .NET, 88 bytes

(a,b,c,d)=>((a=System.Math.Sqrt(2*(a+System.Math.Sqrt(a*a+b*b+c*c+d*d))))/2,b/a,c/a,d/a)

Port of my Java 8 answer, but returns a Tuple instead of a String. I thought that would have been shorter, but unfortunately the Math.Sqrt require a System-import in C# .NET, ending up at 4 bytes longer instead of 10 bytes shorter.. >.>

The lambda declaration looks pretty funny, though:

System.Func<double, double, double, double, (double, double, double, double)> f =

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Perl 6, 49 bytes

{;(*+@^b>>².sum**.5*i).sqrt.&{.re,(@b X/2*.re)}}

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Curried function taking input as f(b,c,d)(a). Returns quaternion as a,(b,c,d).

Explanation

{;                                             }  # Block returning WhateverCode

     @^b>>².sum**.5     # Compute B of quaternion written as q = a + B*u

                        # (length of vector (b,c,d))

  (*+              *i)  # Complex number a + B*i

                      .sqrt  # Square root of complex number

                           .&{                }  # Return

                              .re,  # Real part of square root

                                  (@b X/2*.re)  # b,c,d divided by 2* real part