Solve the following equations:
$6{x^4} + {x^2}{y^2} + 16 = 2x(12x + {y^3}),{\text{ }}{x^2} + xy - {y^2} = 4$

Hint: Try to make given equations in quadratic form. We say an equation is in quadratic form, if it’s in the form of $a{x^2} + bx + c = 0$ .

Complete step-by-step answer:

The given equations are
$

  6{x^4} + {x^2}{y^2} + 16 = 2x(12x + {y^3}) - - - - (1) \\

  {x^2} + xy - {y^2} = 4 - - - - (2) \\

$

Consider equation (2) that is ${x^2} + xy - {y^2} - 4 = 0$. It’s a quadratic equation and we know that if $a{x^2} + bx + c = 0$ is a quadratic equation then, its roots can be written as $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. On comparing equation (2) with the standard form of quadratic equation we’ll get, $a = 1,b = y,c = - {y^2} - 4$. Putting these values back to the root formula:

$\

  x = \dfrac{{ - y \pm \sqrt {{y^2} - 4 \times 1 \times ( - {y^2} - 4)} }}{{2 \times 1}} \\

   \Rightarrow x = \dfrac{{ - y \pm \sqrt {{y^2} + 4({y^2} + 4)} }}{2} \\

   \Rightarrow x = \dfrac{{ - y \pm \sqrt {{y^2} + 4{y^2} + 16} }}{2} \\

   \Rightarrow x = \dfrac{{ - y \pm \sqrt {5{y^2} + 16} }}{2} \\

$

So, the roots of equation (2) are $x = \dfrac{{ - y + \sqrt {5{y^2} + 16} }}{2}{\text{ and }}x = \dfrac{{ - y - \sqrt {5{y^2} + 16} }}{2}$ . But there is a problem in these values. It’s involving y terms which we don’t want. We want to get rid of it. For this we’ll use equation (1). On putting $x = \dfrac{{ - y + \sqrt {5{y^2} + 16} }}{2}$ in the first equation we’ll get $y = 2$. Now let’s substitute this value of y in equation (2) then we’ll get,

$

  {x^2} + xy - {y^2} - 4 = 0 \\

   \Rightarrow {x^2} + 2x - 8 = 0 \\

   \Rightarrow {x^2} + (4 - 2)x - 8 = 0 \\

   \Rightarrow {x^2} + 4x - 2x - 8 = 0 \\

   \Rightarrow x(x + 4) - 2(x + 4) = 0 \\

   \Rightarrow (x - 2)(x + 4) = 0 \\

   \Rightarrow x = 2, - 4 \\

$

And, on substituting $y = - 2$ in equation (2).

$

  {x^2} - 2x - 8 = 0 \\

   \Rightarrow {x^2} - 4x + 2x - 8 = 0 \\

   \Rightarrow x(x - 4) + 2(x - 4) = 0 \\

   \Rightarrow (x - 4)(x + 2) = 0 \\

   \Rightarrow x = 4, - 2 \\

$

And hence the required solutions are,

$
  x = - 2, y = - 2 \\

  x = 4, y = - 2 \\

  x = 2, y = 2 \\

  x = - 4, y = 2 \\
$

Note: Be careful with the calculation part, many students get confused while solving the quadratic equation. Calculation is the crucial point to solve this question.