Lesson 5: Extensions and Applications

Property

To factor a trinomial in two variables of the form $ax^2 + bxy + cy^2$, we use the same methods as for single-variable trinomials. The first and last terms of the trinomial are quadratic, and the middle is a cross-term.
The factored form will look like $(px + qy)(rx + sy)$.

Examples

• To factor $x^2 + 5xy + 6y^2$, we need factors of $6y^2$ that sum to $5xy$. The factors $2y$ and $3y$ work, giving $(x + 2y)(x + 3y)$.

• To factor $a^2 - ab - 12b^2$, we need factors of $-12b^2$ that sum to $-ab$. The factors $-4b$ and $3b$ work, resulting in $(a - 4b)(a + 3b)$.

Property

When a quadratic equation contains a parameter $k$, the discriminant $\Delta = b^2 - 4ac$ determines solution existence. Critical parameter values occur when $\Delta = 0$ (one solution), $\Delta > 0$ (two solutions), or $\Delta < 0$ (no real solutions). Additionally, parameter values that make denominators zero must be excluded from the domain.

Examples

Property

An equation is of quadratic form if a substitution gives us an equation of the form $au^2+bu+c=0$. We look for a relationship where the variable part of the first term is the square of the variable part of the middle term. For example, in $6x^4 - 7x^2 + 2 = 0$, we see $(x^2)^2 = x^4$.

How to solve equations in quadratic form:

1. Identify a substitution that will put the equation in quadratic form.
2. Rewrite the equation with the substitution to put it in quadratic form.
3. Solve the quadratic equation for $u$.
4. Substitute the original variable back into the results, using the substitution.
5. Solve for the original variable.
6. Check the solutions.

Examples

• Solve $x^4 - 10x^2 + 9 = 0$. Let $u=x^2$. This gives $u^2 - 10u + 9 = 0$, so $(u-9)(u-1)=0$. Thus $u=9$ or $u=1$. Substituting back, $x^2=9$ gives $x=\pm3$, and $x^2=1$ gives $x=\pm1$.