"Factorise" a quadratic [duplicate]

Question

When learning to factorise quadratics in the form \$x^2 + ax + b\$, a common technique is to find two numbers, \$p, q\$ such that

$$pq = b \\ p + q = a$$

as, for such numbers, \$x^2 + ax + b = (x + p)(x + q)\$

You are to take two integers \$a, b \in (-\infty, +\infty)\$ and output the two integers \$p, q\$ such that

$$pq = b \\ p + q = a$$

You may take input in any convenient method and you may assume that a solution always exists.

This is code-golf so the shortest code in bytes wins

Test cases

2, -15 -> 5, -3
-22, 85 -> -5, -17
6, -16 -> -2, 8
-8, -240 -> -20, 12
-1, -272 -> 16, -17
17, 16 -> 1, 16
-4, 0 -> 0, -4
3, -54 -> 9, -6
13, 40 -> 8, 5
-29, 198 -> -11, -18
11, -12 -> -1, 12
4, -320 -> 20, -16
4, 4 -> 2, 2

Answer 1

Scala 3, 49 44 bytes

a=>b=>(-b*b to b*b)map(p=>a-p->p)find(_*_==b)

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Takes (a)(b) and returns an Option[(Int, Int)]. It's now a little more inefficient since it goes from \$-b^2\$ to \$b^2\$ instead of \$-|b|\$ to \$|b|\$, including values that \$p\$ and \$q\$ could never be, but it saves 4 bytes.

a => b =>
  (-b*b to b*b)              //Make a range of all possible q's and then some
    map(q => (a - q, q))     //Make a tuple of (p, q)
    find(_ * _ == b)         //Find a pair such that p * q = b

Answer 2

JavaScript (ES7),  37 36 34  31 bytes

Saved 3 bytes thanks to @tsh

Expects (a)(b).

a=>b=>[b=a/2+(a*a/4-b)**.5,a-b]

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Answer 3

Wolfram Language (Mathematica), 23 21 18 bytes

Solve[x x+#2==x#]&

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Answer 4

Jelly, 5 bytes

1ṭÆrN

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Input as a list [b, a].

1ṭ       Append 1 to the input, (resulting in [b, a, 1])
  Ær     find roots of polynomial from little-endian coefficient list,
    N    negate each root.

Answer 5

Retina 0.8.2, 94 bytes

\d+
$*
^(-?)(1*)(,(-)(?<2>\2(1(1)*))|(?<5>(?<6>1)*),)(?<-6>\2)*$(?(6)1)
$1$.2,$4$1$.5
--

-0
0

Try it online! Link includes test cases. Explanation: \$|p|\$ is captured in $2 and \$|q|\$ is captured in $5, with $6 being a helper balancing group used to verify the multiplication.

\d+
$*

Convert to unary.

^(-?)(1*)

Start by matching the sign of \$a\$, then initially assume \$|p|\$ is no greater than \$|a|\$.

(,(-)(?<2>\2(1(1)*))

If \$b\$ is negative, then we actually end up matching \$|a|\$, and choose \$|q|\$ and overwrite $2 with \$|p|=|a|+|q|\$, so \$|a|=|p|+|q|\operatorname{sgn}(b)\$.

|(?<5>(?<6>1)*),)

Otherwise \$b\$ is positive, so overwrite $5 with \$|q|\$ such that \$|a|=|p|+|q|\operatorname{sgn}(b)\$.

(?<-6>\2)*$(?(6)1)

Ensure \$|b|=|pq|\$.

$1$.2,$4$1$.5

Output \$p = |p|\operatorname{sgn}(a)\$ and \$q = |q|\operatorname{sgn}(ab)\$. We then have \$p+q=|p|\operatorname{sgn}(a)+|q|\operatorname{sgn}(ab)=(|p|+|q|\operatorname{sgn}(b))\operatorname{sgn}(a)=|a|\operatorname{sgn}(a)=a\$ while \$pq=|p|\operatorname{sgn}(a)|q|\operatorname{sgn}(ab)=|pq|\operatorname{sgn}(b)=|b|\operatorname{sgn}(b)=b\$ as desired (where \$\operatorname{sgn}(0)=1\$).

--

-0
0

Fix up extraneous -s.

Answer 6

Charcoal, 25 bytes

ＮθＮηＩ⊟ΦＥ…·±↔η↔η⟦ι⁻θι⟧⁼ηΠι

Try it online! Link is to verbose version of code. Same idea as @user's Scala solution. Explanation:

ＮθＮη

Input \$a\$ and \$b\$.

Ｅ…·±↔η↔η

Loop \$p\$ over \$[-|b|,|b|]\$.

⟦ι⁻θι⟧

For each \$p\$ create a list \$(p,q)\$ where \$q=a-p\$.

Ｉ⊟Φ...⁼ηΠι

Output the one with the highest \$p\$ where \$pq=b\$.

Answer 7

Octave, 17 bytes

@(x)-roots([1 x])

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You simply need to solve \$x^2 + ax + b = (x + p)(x + q) = 0\$, and \$x_1=-p\$, \$x_2=-q\$.

Answer 8

Python 3.8, ^{42 39} 37 bytes

Saved 2 bytes thanks to Danis!!!

lambda a,b:[c:=a/2+(a*a/4-b)**.5,a-c]

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Port of Arnauld's JavaScript answer.

Answer 9

05AB1E, 17 16 bytes

(Dn4I*-t‚DOsÆ);(

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Why oh why aren't there any built-ins for this? This solves the quadratic equation to get the two roots. -1 thanks to @ovs

Answer 10

Brachylog, 8 bytes

~+ʰ⟨≜×⟩Ċ

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Takes input as a list [a,b].

  ʰ         For the first element of the input,
~+          get a list of integers which sum to it
    ≜       (in order of least absolute value of the first element).
   ⟨ ×⟩     Its product is the second element,
       Ċ    its length is 2,
            and it's the output.

I'm not entirely sure why this needs ≜, but it comes at no cost because the "sandwich" construct ⟨~+×⟩ won't parse without braces around ~+ anyhow (and both ⟨{~+}×⟩Ċ and ~+ʰ⟨≡×⟩Ċ would still come out to 8).

Answer 11

Wolfram Language (Mathematica), 28 bytes 24 bytes

thanks to att

#&@@Solve[#-b==a==#2/b]&

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Answer 12

dc, 24 bytes

Input is a line of two numbers from STDIN. Output is to STDOUT.

?rdsad*r4*-vla+2/dlar-rf

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Explanation

?                        # Read a line of input
 r                       # Swap the two numbers
  dsa                    # Store to `a` w/o popping
     d*                  # Square the 1st input
       r                 # Swap up the second input
        4*               # Multiply by four
          -              # Second-to-top - TOS
           v             # Square root
            la+          # Add by `a`
               2/        # Halve
                 d       # Duplicate
                  lar-   # Push `a` - TOS
                      r  # Swap the top 2 items
                       f # Print entire stack

Answer 13

Python 3.8 (pre-release), 37 bytes

lambda a,b:[p:=a/2+(a*a/4-b)**.5,a-p]

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Python 2, 42 bytes

def f(a,b):p=a/2+(a*a/4-b)**.5;print p,a-p

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Takes input as float, and prints the output.

Answer 14

Husk, 12 bytes

ḟo=¹ΠmSe`-²ṡ

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Port of user's Scala answer.

           ṡ    # range of integers from -arg1 to +arg1
     m          # map over each of them
      S         # combining itself
       e        # into a 2-element list with
        `-²     # arg2 minus this;
ḟ               # now find the first element that satisfies
 o              # combine 2 functions:
    Π           # product
  =¹            # equals arg1

Answer 15

R, 36 35 bytes

function(b,c)b/2+-.5:1*(b^2-4*c)^.5

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Uses the Quadratic formula with a=1.

Answer 16

C (gcc) -lm, 45 bytes

f(int*a){*a-=a[1]=(*a+sqrt(*a**a-4*a[1]))/2;}

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Modifies an input array of 2 ints.

---

C (gcc), 53 bytes

i;f(int*a){for(i=*a/2;(*a-i)*i-a[1];i--);*a-=a[1]=i;}

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Doesn't use sqrt.

Answer 17

APL (Dyalog Unicode), 28 26 bytes

-2 thanks to Razetime.

{⍺-⎕←(⍺÷2)+.5*⍨⍵-⍨⍺×⍺÷4}

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Answer 18

APL (Dyalog Extended), 19 bytes

-2 thanks to @Bubbler

÷∘2⍛{⍺(+,-)√⍵-⍨⍺*2}

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Answer 19

Haskell, 46 39 bytes

a!b=[(k,b-k)|k<-[-a*a..],k*b-k*k==a]!!0

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Kind of boring, but short.

Answer 20

convey, 113 bytes

v<1
>>>v 0
{?>*"-"v
4*  "vvv
 v  >*vv
v#<<<<vv
-<    vv
*%2   v>v
.1    v}>v
v     v%2v}%2
">>>>>-^>+>^
>>>>>>>>^

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Ungolfed, mostly. Uses quadratic formula.

Answer 21

Julia 1.0, 34 bytes

(a,b)->(d=(a*a-4b)^.5;[a+d,a-d]/2)

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Answer 22

MathGolf, 13 bytes

_½\²¼k-√+`-αÞ

I/O as floats.

Port of @Arnauld's JavaScript answer.

Try it online.

I have the feeling this can be a byte shorter with some smart stack manipulations, but I've been unable to find anything yet. An equal-bytes alternative could be ²¼k-√`;½+`-αÞ.

Explanation:

_             # Duplicate the first (implicit) input-float
 ½            # Halve it
  \           # Swap so the first input is at the top of the stack again
   ²          # Square it
    ¼         # Divide it by 4
     k-       # Subtract the second input-float
       √      # Take the square-root of that
        +     # Add it to the halved first input that's still on the stack
         `    # Duplicate the top two values (since the stack only contains a single
              # item, this will first add the first input-float implicitly as leading
              # item, and then duplicate both items)
          -   # Subtract the top two items from one another
           α  # Pair the top two items
            Þ # Only leave the top item of the stack, and discard everything else
              # (after which the entire stack is output implicitly as result)

Answer 23

Ruby, 41 37 35 bytes

A port of Arnauld's answer in Ruby!

->a,b{[b=a/2r+(a*a/4r-b)**0.5,a-b]}

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Answer 24

05AB1E, 10 9 bytes

D(Ÿʒ-y*¹Q

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Takes input as

b
a

In case \$p=q\$ it returns only a single value. In case that's not allowed, it's +5 bytes for appending }Ðgi«

Explanation:

D              duplicate
)              negate
Ÿ              range, so [-b,.. , b]
ʒ              filter
 -             minus the implicit a
  y            the current number
   *           multiply
    ¹          the first item from the input history - b
     Q         is equals

Answer 25

Vyxal, 3 bytes

∆QN

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Solve for the roots and then negate

Answer 26

Whispers v2, 110 92 bytes

Edit: -18 bytes (within less than 2 minutes of submission!) thanks to caird coinheringaahing and the ± operator

> Input
> Input
> 2
> 4
>> 1*3
>> 2⋅4
>> 5-6
>> √7
>> 8÷3
>> 1÷3
>> 10±9
>> Output 11

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My first Whispers program.
Uses the Quadratic formula with a=1.

Answer 27

Pari/GP, 33 bytes

f(a,b)=round(polroots(x^2+a*x+b))

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Answer 28

Haskell, 39 bytes

f a b=let c=a/2+(a*a/4-b)**0.5in[c,a-c]

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