Solving disguised quadratics

​​In a nutshell

Disguised quadratics are equations that are not strictly quadratics, but can be expressed and solved as a quadratic. You know many methods to solve quadratic equations. So, it is helpful to turn a non-quadratic equation into a quadratic equation in order to solve. Disguised quadratics can be turned into quadratics with a substitution.

Spotting the disguise

Quadratics equations are given in the form 

​$$ax^2+bx+c=0$$​​

where $$a$$, $$b$$ and $$c$$ are constants. This is referred to as a "quadratic in $$x$$". The following is a "quadratic in $$p$$":

​$$ap^2+bp+c=0$$​​

You can replace $$x$$ and you now have either a quadratic, or something that can be solved like a quadratic. Consider replacing the $$p$$ now with $$x^2$$:

​$$a(x^2)^2+b(x^2)+c=0$$​​

This is the same as

​$$ax^4+bx^2+c=0$$​​

This is an example of a disguised quadratic. It is not a quadratic, in fact it is a quartic, but it can be solved like a quadratic equation.

Example 1

Solve the equation:

$$x^4-5x^2+6=0$$​​

One step you can take here is to make a substitution to make the equation look like a quadratic. For example, substitute $$x^2$$ for $$t$$:

$$t^2-5t+6=0$$ where $$t=x^2$$​​

Now solve for $$t$$:

$$(t-2)(t-3)=0$$​​

So $$t=2$$ or $$t=3$$. But you have to find $$x$$. So, the solutions will come instead from the equations:

 $$x^2=2$$, $$x^2=3$$​

Solving each of these gives:

$$\underline{x=\pm\sqrt2,\pm\sqrt{3}}$$​​

These are the four solutions to the equation $$x^4-5x^2+6=0$$.

Example 2

Solve the equation:

$$x-9\sqrt{x}+20=0$$​​

This is a disguised quadratic and can be shown more clearly with a substitution of $$\sqrt{x}=y$$:

$$y^2-9y+20=0$$​​

Factorise this:

$$(y-5)(y-4)=0$$​​

So $$y=5$$ or $$y=4$$. Therefore:

$$\sqrt x=5$$, $$\sqrt x=4$$

Square these to find the possible values of $$x$$. So the solutions of $$x+9\sqrt{x}+20=0$$ are:

$$\underline{x=25,x=16}$$​​

Note: Always check your solutions with the original equation. Inserting these two values shows that they do both satisfy the equation.

Example 3

Solve the equation:

$$y^4+5y^2-36=0$$​​

This is a disguised quadratic because it has a constant term, then the powers involved have a constant difference: it may help to consider the constant term as $$-36y^0$$, since now you see the powers jump from $$0$$ to $$2$$ to $$4$$. When the jumps are constant like this, you have a disguised quadratic.

The substitution is $$x=y^2$$:

$$x^2+5x-36=0$$​​

Factorising this gives 

$$(x+9)(x-4)=0$$​​

Hence $$x=-9$$ or $$x=4$$. By reintroducing the substitution, you have:

$$y^2=-9$$, $$y^2=4$$

But a squared number cannot equal a negative, so you will only get solutions from $$y^2=4$$. Thus the only solutions to the original equation are:

$$\underline{y=\pm2}$$​​​