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Solving disguised quadratics

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Summary

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:

$\underline{y=\pm2}$ 

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Unit 1

Factorising quadratics

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The quadratic formula - Higher

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