We have already solved quadratic equations using several algebraic methods. But what if we are really only interested in what types of solutions exist? Are they real and rational? Real and irrational? Imaginary? Are there always two solutions or can there be just one solution?
If the quadratic equation is written in standard form \(ax^2-bx+c=0\), we can use the discriminant, which is the number UNDER the square root sign in the quadratic formula, to answer these questions. Think about it--we can get positive radicands, negative radicands or a radicand of zero (if you are wondering, a radicand is the number under the radical sign). Since the square root of a positive number is real, we would have two real solutions in this case; since the square root of a negative number is imaginary, we would have two imaginary solutions in this situation; and since the square root of zero equals zero, you would get one real solution.
Let’s formalize this with a formula.
If the quadratic equation is written in standard form \(ax^2-bx+c=0\), we can use the discriminant, which is the number UNDER the square root sign in the quadratic formula, to answer these questions. Think about it--we can get positive radicands, negative radicands or a radicand of zero (if you are wondering, a radicand is the number under the radical sign). Since the square root of a positive number is real, we would have two real solutions in this case; since the square root of a negative number is imaginary, we would have two imaginary solutions in this situation; and since the square root of zero equals zero, you would get one real solution.
Let’s formalize this with a formula.
The quadratic formula is \(x=\large\frac{-b\pm\sqrt{b^2-4ac}}{2a}\), so the discriminant is \(D=b^2-4ac\).
The table below summarizes how many solutions will exist for different discriminant values.
Here is the other great thing about the discriminant--if it is equal to a perfect square or zero, you know the quadratic is factorable! So if you are not sure which solving method you should use, this can be an easy mental math method to determine if factoring is an option.
Example 1: How many and what types of solutions exist for the quadratic equation \(3x^2+6=5x\)?
\(\begin{align}&3x^2-5x+6=0\ & \ &\text{1) Write the equation in standard form.}\\&a=3,\ b=-5,\ c=6\ & \ &\text{2) Identify the values of a, b and c.}\\&b^2-4ac=\left(-5\right)^2-4\left(3\right)\left(6\right)\ & \ &\text{3) Substitute the values in the formula.}\\&b^2-4ac=25-72\ & \ &\text{4) Calculate the discriminant.}\\&b^2-4ac=-47\end{align}\)
Since the discriminant is negative, there are two complex solutions.
Example 2: How many and what types of solutions exist for the quadratic equation \(-4x^2+2x+1\)?
\(b^2-4ac=2^2-4\left(-4\right)\left(1\right)=4+16=20\)
Since the discriminant is positive, and \(20\) is not a perfect square, there are two real, irrational solutions.
Example 3: How many and what types of solutions exist for the quadratic equation \(x^2-6x+9=0\)?
Example 2: How many and what types of solutions exist for the quadratic equation \(-4x^2+2x+1\)?
\(b^2-4ac=2^2-4\left(-4\right)\left(1\right)=4+16=20\)
Since the discriminant is positive, and \(20\) is not a perfect square, there are two real, irrational solutions.
Example 3: How many and what types of solutions exist for the quadratic equation \(x^2-6x+9=0\)?
Example 4: How many and what types of solutions exist for the quadratic equation \(6x^2+x-2=0\)?
How can you determine the number of solutions if the equation is written in vertex form?
You cannot use the discriminant if the quadratic function is written in vertex form. But you can determine if there is one real, two real or two imaginary solutions by making a quick sketch--plot the vertex and determine if the parabola opens up or down based on whether the leading coefficient is positive (up) or negative (down). If you recall, real solutions occur at the x-intercepts. If there are two x-intercepts there are two real solutions (although you cannot determine if these solutions are rational or irrational), if there is one x-intercept (the vertex is on the x-axis) there is one real solution and if there are no x-intercepts, there are two complex solutions.
Example 5: Which of the following quadratic functions have two real solutions? Select all that apply.
a) \(f\left(x\right)=-3\left(x-1\right)^2+5\)
b) \(f\left(x\right)=3\left(x+4\right)^2+7\)
c) \(f\left(x\right)=-5x^2+4x+1\)
d) \(f\left(x\right)=4x^2+20x+25\)