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State 2 values of $x$ for which the value of $3x^2+4x-14$ is a perfect square.

I can't seem to factor it and I'm really lost. Does using the quadratic equation for it work? Help I need it for grade 10 quadratics test.

Maadhav
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Ajwad Kabir
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  • The quadratic formula is indeed a way to solve this. You want this polynomial to be a perfect square at $x$, so you want $$3x^2+4x-14=n^2\iff 3x^2+4x-(14+n^2)=0$$for some $n\in\Bbb N$. – Dave Dec 02 '17 at 02:20
  • Note, if you consider $0$ a perfect square, you just find the two roots of this polynomial and you're done. – Dave Dec 02 '17 at 02:24

4 Answers4

1

$$(1)\quad3x^2+4x-14=a^2$$ $$\implies 3x^2+4x-(14+a^2)=0\implies x=\dfrac{-4\pm\sqrt{16-12(-14-a^2)}}{6}$$

This does not need to be an integer. This says that for $(1)$ to be true, just choose an $a$ here and it will give you an $x$ value that satisfies. For example, $a=1$: $$x=\dfrac{-4\pm\sqrt{16-12(-15)}}{6}=\dfrac{-4\pm\sqrt{196}}{6}=\frac{10}{6},-3$$ Plugging either one of these values into the original equation will give you $a=1$.

user12345
  • 2,149
1

$$3x^2+4x-14=n^2\\x=\frac{-4\pm\sqrt{16+12(14+n^2)}}{6}=-\frac23\pm\frac13\sqrt{46+3n^2}$$

Setting $n=1$ gives a nice value for the square root: $7$. In this case, you get $x=-3,\frac53$ as two values of $x$ for which your quadratic gives a square number, namely $1$, as the value.

This choice of $n$ is not unique, you could have chosen any natural number. This is just the simplest one.

John Doe
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I AM ASSUMING THAT THE QUESTION IS ASKING FOR $x$ an integer.

I finished up. There are a two-ended sequence of positive $x$ values (integers) that work, the linear recurrence can be used in either direction, $$ x_{n+2} = 14 x_{n+1} - x_n + 8, $$ all positive are $$ ..., 8853, 635, 45,3, 5, 75, 1053, 14675,...$$ The double ended sequence of negative $x$ values is $$ ... -33043, -2373, -171, -13, -3, -21, -283, -3933, -54771,... $$ Together these are ALL integer $x$ values.

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This becomes a Pell type equation, $$ (3x+2)^2 - 3 y^2 = 46. $$ By nature, there are answers both with positive $x$ and negative $x.$

If you prefer positive $x,$ there are two infinite sequences. One is $$ 3, 45, 635, 8853, ... $$ with $$ x_{n+2} = 14 x_{n+1} - x_n + 8. $$ For this one the results of $3x^2 + 4 x - 14$ are $y^2,$ with $y$ in $$ 5, 79, 1101, 15335,... $$ and $$ y_{n+2} = 14 y_{n+1} - y_n . $$

The other positive sequence is $$ 5, 75, 1053, 14675,... $$ also with

$$ x_{n+2} = 14 x_{n+1} - x_n + 8. $$ For this one the results of $3x^2 + 4 x - 14$ are $y^2,$ with $y$ in $$ 9, 131, 1825, 25419... $$ and $$ y_{n+2} = 14 y_{n+1} - y_n . $$

=====================================================================================

jagy@phobeusjunior:~$ ./Pell_Target_Fundamental
  Automorphism matrix:  
    2   3
    1   2
  Automorphism backwards:  
    2   -3
    -1   2

  2^2 - 3 1^2 = 1

 w^2 - 3 v^2 = 46

Sat Dec  2 10:20:49 PST 2017

w:  7  v:  1  SEED   KEEP +- 
w:  11  v:  5  SEED   BACK ONE STEP  7 ,  -1
w:  17  v:  9
w:  37  v:  21
w:  61  v:  35
w:  137  v:  79
w:  227  v:  131
w:  511  v:  295
w:  847  v:  489
w:  1907  v:  1101
w:  3161  v:  1825
w:  7117  v:  4109
w:  11797  v:  6811
w:  26561  v:  15335
w:  44027  v:  25419
w:  99127  v:  57231
w:  164311  v:  94865
w:  369947  v:  213589
w:  613217  v:  354041
w:  1380661  v:  797125
w:  2288557  v:  1321299
w:  5152697  v:  2974911
w:  8541011  v:  4931155
w:  19230127  v:  11102519
w:  31875487  v:  18403321

Sat Dec  2 10:21:49 PST 2017

 w^2 - 3 v^2 = 46

In this one I told it to show the positive $x$ when the left hand number in the line is $2 \pmod 3,$ otherwise give $x$ negative.

Sat Dec  2 15:38:39 PST 2017
                   7                  -3                   1
                  11                   3                   5
                  17                   5                   9
                  37                 -13                  21
                  61                 -21                  35
                 137                  45                  79
                 227                  75                 131
                 511                -171                 295
                 847                -283                 489
                1907                 635                1101
                3161                1053                1825
                7117               -2373                4109
               11797               -3933                6811
               26561                8853               15335
               44027               14675               25419
               99127              -33043               57231
              164311              -54771               94865
              369947              123315              213589
              613217              204405              354041
             1380661             -460221              797125
             2288557             -762853             1321299
             5152697             1717565             2974911
             8541011             2847003             4931155
            19230127            -6410043            11102519
            31875487           -10625163            18403321
Sat Dec  2 15:39:02 PST 2017
Will Jagy
  • 139,541
-1

3x^2+4x-14=0. Let q and w be 2 roots of the equation. q+w=4,qw=-42 (q-w)^2=(q+w)^2-4qw So q-w=√184=2√46 On solving q+w=4and q-w=2√46 We get w=q={4+2√46}÷\2 Now put the value of w andq and solve it.on verifying q+w is not equal with 4 . slight deviations will come as we have taken imaginary roots.

Swayam Pattanaik
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