An equation is a quadratic equation if it can be written in the form:
\(ax^2+bx+c=0\),
where \(a,b,c\) are known values, \(a \ne 1\), and \(x\) is some unknown variable. It has degree of 2 since the quadratic polynomial has degree 2 (i.e. highest exponent of all monomials in the polynomial is 2: \(x^2\)).
Recall the methods we can use to solve quadratic equations such as factoring or using the quadratic formula (review these on the Solving Quadratic Equations page). These only work for solving quadratic equations, but what if we wanted to solve equations of higher degrees (i.e. degree 3 or higher)?
To solve higher degree equations, we can use substitution to convert the given equation into a quadratic equation, then solve the quadratic equation to determine the solutions to the original equation.
For example, suppose we have the equation:
\(ax^4+bx^2+c=0\)
If we let \(z=x^2\), then substitute this into the original equation, we can rewrite it as:
\(a(x^2)^2+b(x^2)+c=0\)
\(\Rightarrow az^2+bz+c=0\),
which is a quadratic equation that we can solve (by factoring or using the quadratic equation).
Then after solving, we can set the solutions for \(z\) equal to \(x^2\), then solve for \(x\).
Tip: Don't forget to find the solutions of the original equation after solving the rewritten equation!
Solution:
Since we want to rewrite this equation as a quadratic equation, we use substitution by letting \(z=x^3\). So the equation becomes:
\((x^3)^2+5(x^3)+6=0\)
\(\Rightarrow z^2+5z+6=0\)
We can now solve this quadratic equation by factoring, giving us:
\((z+2)(z+3)=0\)
\(\Rightarrow z=-2\) or \(z=-3\)
Finally, we solve for \(x\) using \(z=x^3\):
\(z=-2=x^3\) or \(z=-3=x^3\)
\(\Rightarrow\) \(x=\sqrt[3]{-2}\) or \(x=\sqrt[3]{-3}\)
We can also use this substitution method to solve other types of equations (not only ones involving polynomials), seen in this following example:
Solution:
We can rewrite the rational expression on the left side by substituting with \(z=x^2\):
\(\frac{(x^2)^2+(x^2)-20}{(x^2)+5}=0\)
\(\Rightarrow \frac{z^2+z-20}{z+5}=0\)
Now we can factor the numerator to simplify the rational expression, making note of any variable restrictions (since the denominator cannot equal 0):
\(\frac{\cancel{(z+5)}(z-4)}{\cancel{(z+5)}}=0, \ z \ne 5\)
\(\Rightarrow z-4=0\)
\(\Rightarrow z=4\)
Finally, we solve for \(x\) using \(z=x^2\):
\(z=4=x^2\)
\(\Rightarrow\) \(z=2\) or \(z=-2\)
Note: Recall that there are two solutions to \(x^2=a\) (for any value \(a>0\), if there are no restrictions on the variable \(x\)):
\(x=+\sqrt{a}\) or \(x=-\sqrt{a}\)