$x^2-7x+m=0$, $x_1^2+4 x_2^2=68$, $m$=? $x_1, x_2$ are solutions to the equation and  are whole numbers
$x^2-7x+m$
I dont see how to represent $x_1^2+4 x_2^2=68$ only with $x_1 + x_2$ and  $x_1 x_2$.I could try to use the fact that $x_1, x_2$ are whole numbers and try to guess $m$
$$x_{1,2} = \dfrac{7\pm\sqrt{49-4m}}{2}$$
But i think that is not the proper way to solve this
 A: Hint: If so then $$x_1+x_2=7$$ and $$x_1x_2=m$$
With $$x_2=7-x_1$$ we get
$$x_1^2+4(7-x_1)^2=68$$ so $$x_1=8$$
A: $x^2-7x+m=0\implies x^2=7x-m$ so $$68=x_1^2+4 x_2^2= 7x_1-m+4(7x_2-m)=7(x_1+4x_2)-5m$$ 
so $$ 68 = 7\cdot 7+21x_2-5m\implies x_2= {19+5m \over 21}$$
So if we plug this in starting equation we get $$\Big({19+5m \over 21}\Big)^2-7{19+5m \over 21} +m=0\implies m={104\pm 504\over 50}$$
So $m= {304\over 25}$ or $m= -8$.
A: You can try to express everything in terms of one root. 
I'll call the roots $y$ and $z$ - which both satisfy the original equation, so $y+z=7$ Now put $y=7-z$ so you get $(7-z)^2+4z^2=68$
This gives $49-14z+5z^2=68$ or $5z^2-14z-19=0$ and this factorises as $(5z-19)(z+1)=0$
$z$ is an integer. (this also informs the fact that a factorisation will be possible)
A: 1)
If the two solutions to $ax^2 + bx + c = 0$ are $\frac {-b\pm \sqrt{b^2 - 4ac}}{2a}$ then the sum of these solutions is $\frac {-b+ \sqrt{b^2 - 4ac}}{2a}+\frac {-b- \sqrt{b^2 - 4ac}}{2a} =-\frac ba$.
And the product is $(\frac {-b+\sqrt{b^2 - 4ac}}{2a})(\frac {-b-\sqrt{b^2 - 4ac}}{2a}) = \frac {b^2 - (b^2 -4ac)}{4a^2} = \frac ca$.
So if $x_1, x_2$ are the solutions to $x^2 -7x + m = 0$ then $x_1 + x_2 = 7$ and $x_1x_2 = m$.
2)
Forget all that.
If $x_1$ and $x_2$ are whole numbers there are only finitely many pairs where $x_1^2 + 4x_2^2= 68$.
$4x_2^2 \le x_1^2 + 4x_2^2 = 68$ so $x_2^2 \le 17$ so $|x_2| \le 4$.
Just test all $|x_2|$ from $0$ to $4$.
Or if you feel clever reason that:
$4x_2^2$ and $68$ are even so $x_1^2$ is even so $x_1$ is even.  
If we let $x_1=2c$ then $4c^2  + 4x_2^2 = 68$ or $c^2 + x_2^2 = 17$.
Trial and error give us $0 + x_2^2 = 17$ is impossible $1 + 4^2 = 17$ is possible.  $4 + x_2^2 = 17$ is impossible.  $9 + x_2^2 =17$ is impossible and $4 + 1^2 = 17$ are possible and nothing else is.
So $(c, x_2) = \{(\pm 1, \pm 4), (\pm 4, \pm 1)\}$ and
$(x_1,x_2) = (2c,x_2) = \{(\pm 2,\pm 4)(\pm 8, \pm 1)\}$ are the only possible answers.
.....
Now back to 1)
But if $x_1 + x_2 = 7$ the only possible options are $x_1 = 8$ and $x_2 =-1$.
So $x_1x_2 = -8$.
And so as $m= x_1x_2 = -8$.
