Section 4.6
The Discriminant
The discriminant of a quadratic equation is the expression $$b^2-4ac$$ that appears under the square root in the Quadratic Formula. The value of the discriminant allows us to determine the nature of the solutions of the quadratic equation. We start with a quadratic equation in standard form $$ax^2+bx+c=0$$ and note that there are three possibilities.
- If $b^2-4ac>0$, then the quadratic equation has two real solutions.
- If $b^2-4ac=0$, then the quadratic equation has exactly one real solution.
- If $b^2-4ac<0$, then the quadratic equation has no real solutions.
Examples:
a. $x^2-3x-7=0$ has the discriminant $(-3)^2-4\cdot 1\cdot 1\cdot (-7)=9+28$. Since this is a positive number, the equation has two real solutions. They are $x=\frac{3\pm \sqrt{37}}{2}$.
b. The discriminant of the equation $x^2+12x+36=0$ is $12^2-4\cdot 1\cdot 36=144-144=0$. Since this is zero, the quadratic equation has just one solution: $x=-6$.
c. The discriminant of the equation $2x^2+x+5=0$ is $1^2-4\cdot 1\cdot 5=1-20$. Since this is a negative number, the quadratic equation has no solutions.
In addition to this, assuming that our quadratic equation has integer coefficients (that is, if all three of $a$, $b$, and $c$ are integers), then
- if the discriminant is a perfect square, the solutions of the equation will be rational numbers; and
- if the discriminant is not a perfect square, the solutions of the equation will be irrational numbers.
d. $5x^2-6x+1=0$ has the discriminant $(-6)^2-4\cdot 5 \cdot 1=36-20=16$. Since 16 is a perfect square, this quadratic has rational solutions. The solutions are $x=1$ and $x=\frac{1}{5}$.
e. The discriminant of the equation $5x^2+8x+2=0$ is $8^2-4\cdot 5\cdot 2=64-40=24$. Since 24 is not a perfect square, the solutions of this quadratic are irrational. They are $x=\frac{-8\pm\sqrt{24}}{10}=\frac{-8\pm 2\sqrt{6}}{10}=\frac{2(-4\pm\sqrt{6})}{10}=\frac{-4\pm\sqrt{6}}{5}$.
Discriminant of quadratic equations: Discriminant of Quadratic Equations
Discriminant for types of solutions for a quadratic: Discriminant for Types of Solutions for a Quadratic
Exercises
Compute the discriminant of each of the following. Use the resulting value to find out how many solutions the quadratic has and whether they are rational or irrational.\begin{array}{l l l}
1. \;\;{{x}^{2}}-7x-60=0\;\;& 2. \;\;{{x}^{2}}-7x+20=0\;\;& 3. \;\;4{{x}^{2}}-12x+9=0\;\; \\
4. \;\;2{{x}^{2}}-3x+4=0\;\;& 5. \;\;2{{x}^{2}}-3x+1=0\;\;& 6. \;\;{{x}^{2}}-7x+10=0\;\; \\
7. \;\;5{{x}^{2}}-7x+1=0\;\;& 8. \;\;[5{{x}^{2}}-7x+2=0\;\;& 9. \;\;5{{x}^{2}}-7x+3=0\;\; \\
10. \;\;x(x+1)=3\;\;& 11. \;\;4{{x}^{2}}-28x+49=0\;\;& 12. \;\;2{{x}^{2}}-4x-12={{x}^{2}}-x-2\;\;
\end{array}
For what value(s) of $K$ will each of the following equations have exactly one solution?
\begin{array}{l l}
13. \;\;{{x}^{2}}+8x+K=0\;\;& 14. \;\;{{x}^{2}}+Kx+K=0\;\;& 15. \;\;2{{x}^{2}}+Kx+3=0\;\; \\
16. \;\;2K{{x}^{2}}+Kx+5=0\;\;& 17. \;\;Kt-5{{t}^{2}}=60\;\;& 18. \;\;Kt-5{{t}^{2}}=100\;\;
\end{array}