Equations Which Are Quadratic in Form

Sometimes an equation is not quadratic as is, but becomes quadratic if you make a substitution. Then you can solve the resulting quadratic, and get solutions to the original equation by using the substitution equation. Equations of this type are said to be quadratic in form.


Example. Solve $x^6 - 3 x^3 - 4 = 0$ . (Complex solutions are allowed.)

Write the equation as

$$(x^3)^2 - 3 x^3 = 4 = 0.$$

The equation is actually quadratic; you can see this more clearly by substituting y for $x^3$ :

$$y^2 - 3 y - 4 = 0.$$

Factor and solve:

$$\matrix{& & y^2 - 3 y - 4 = 0 & & \cr & & (y - 4)(y + 1) = 0 & & \cr & \swarrow & & \searrow & \cr y - 4 = 0 & & & & y + 1 = 0 \cr y = 4 & & & & y = -1 \cr}$$

$y = 4$ gives $x^3 =
   4$ , or $x = \root 3 \of 4$ .

$y = -1$ gives $x^3 =
   -1$ , or $x = \root 3 \of {-1} = -1$ .

The solutions are $x = \root
   3 \of 4$ and $x = -1$ .


Example. Solve $(x + 3)^2 - 4(x + 3) - 5 = 0$ . (Complex solutions are allowed.)

Let $y = x + 3$ . The equation becomes

$$y^2 - 4 y - 5 = 0.$$

Factor and solve:

$$\matrix{& & y^2 - 4 y - 5 = 0 & & \cr & & (y - 5)(y + 1) = 0 & & \cr & \swarrow & & \searrow & \cr y - 5 = 0 & & & & y + 1 = 0 \cr y = 5 & & & & y = -1 \cr}$$

$y = 5$ gives $x + 3
   = 5$ , or $x = 2$ .

$y = -1$ gives $x + 3
   = -1$ , or $x = -4$ .

The solutions are $x = 2$ or $x = -4$ .


Example. Solve the equation $x^{-2} - 3 x^{-1} - 4 = 0$ . (Complex solutions are allowed.)

Write the equation as

$$(x^{-1})^2 - 3 x^{-1} - 4 = 0.$$

Let $y = x^{-1}$ . Then

$$\matrix{ & & y^2 - 3 y - 4 = 0 & & \cr & & (y - 4)(y + 1) = 0 & & \cr & \swarrow & & \searrow & \cr y = 4 & & & & y = -1 \cr \noalign{\vskip2pt} x^{-1} = 4 & & & & x^{-1} = -1 \cr \noalign{\vskip2pt} \dfrac{1}{x} = 4 & & & & \dfrac{1}{x} = -1 \cr \noalign{\vskip2pt} x = \dfrac{1}{4} & & & & x = -1 \quad\halmos \cr}$$


Example. Solve the equation $x^{2/3} - x^{1/3} - 2 = 0$ . (Complex solutions are allowed.)

Write the equation as

$$(x^{1/3})^2 - x^{1/3} - 2 = 0.$$

Let $y = x^{1/3}$ . Then

$$\matrix{ & & y^2 - y - 2 = 0 & & \cr & & (y - 2)(y + 1) = 0 & & \cr & \swarrow & & \searrow & \cr y = 2 & & & & y = -1 \cr \noalign{\vskip2pt} x^{1/3} = 4 & & & & x^{1/3} = -1 \cr (x^{1/3})^3 = 4^3 & & & & (x^{1/3})^3 = (-1)^3 \cr x = 64 & & & & x = -1 \quad\halmos \cr}$$


Example. Solve the equation $(2 x + 3)^2 + 2(2 x + 3) - 3 =
   0$ . (Complex solutions are allowed.)

Let $y = 2 x + 3$ . Then

$$\matrix{ & & y^2 + 2 y - 3 = 0 & & \cr & & (y - 1)(y + 3) = 0 & & \cr & \swarrow & & \searrow & \cr y = 1 & & & & y = -3 \cr \noalign{\vskip2pt} 2 x + 3 = 1 & & & & 2 x + 3 = -3 \cr 2 x = -2 & & & & 2 x = -6 \cr x = -1 & & & & x = -3 \quad\halmos \cr}$$


Example. Solve $x^4 - 2 x^2 - 3 = 0$ . (Complex solutions are allowed.)

Write the equation as

$$(x^2)^2 - 2 x^2 - 3 = 0.$$

Let $y = x^2$ . The equation becomes

$$y^2 - 2 y - 3 = 0.$$

Factor and solve:

$$\matrix{& & y^2 - 2 y - 3 = 0 & & \cr & & (y - 3)(y + 1) = 0 & & \cr & \swarrow & & \searrow & \cr y - 3 = 0 & & & & y + 1 = 0 \cr y = 3 & & & & y = -1 \cr}$$

$y = 3$ gives $x^2 =
   3$ , or $x = \pm\sqrt{3}$ .

$y = -1$ gives $x =
   -1$ , or $x = \pm i$ .

The solutions are $x =
   \pm\sqrt{3}$ and $x = \pm i$ .

Note: If the problem had asked for only real solutions, then the only solutions are $x =
   \pm \sqrt{3}$ .



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