Let \(a\), \(b\), and \(c\) be real numbers with \(a\neq 0\).
The discriminant of the quadratic equation \(ax^2+bx+c = 0\) is defined as the number \(b^2-4\cdot a \cdot c\).
The reason for introducing the discriminant (hereafter denoted by the letter \(D\)) is that we can now formulate a rule for how many real solutions the quadratic equation has and, if solutions exist, which exact values they have.
The formula below, which directly gives the solutions and their number, is called the quadratic formula.
The quadratic equation \(ax^2+bx+c = 0\) with unknown \(x\) and discriminant \(D=b^2-4ac\) has:
- two real solutions if \(D\gt 0\), namely \(x=\dfrac{-b - \sqrt{D}}{2a}\) and \(x=\dfrac{-b+ \sqrt{D}}{2a}\).
- exactly one real solution if \(D=0\), namely \(x=-\dfrac{b}{2a}\).
- no real solutions if \(D\lt 0\).
Start with the quadratic equation\(ax^2+bx+c=0\) with \(a\neq 0\). Multiply both sides by \(4a\) to obtain \[4a^2+4abx+4ac=0\] Completing the square leads to \[(2ax+b)^2 -b^2+4ac=0\] This can be rewritten as \[(2ax+b)^2-b^2-4ac\]
If \(b^2-4ac\gt 0\) , then we can take the square root on both sides; these two roots are up to a sign equal to each other: \[x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\] So there are two real solutions, namely \[x=\frac{-b - \sqrt{b^2-4ac}}{2a}\quad\text{and}\quad x=\dfrac{-b+ \sqrt{b^2-4ac}}{2a}\]
If \(b^2-4ac= 0\) , then the right-hand side, and thus also the left-hand side, is equal to \(0\), from which it follows that \[x=\dfrac{b}{2a}\] is the only real solution.
If \(b^2-4ac\lt 0\) , then the right-hand side is negative, but the left-hand side is not; so there are no real solutions.
The two solutions in the first case are often written together, by making use of the \(\pm\) notation; so \[x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}\]
Use the quadratic formula to compute the exact real solution of the quadratic equation \[3\, x^2-4\, x+1=0\]
The number of solutions of the quadratic equation \[ax^2+bx+c=0\] depends on the discriminant \(D=b^2-4ac\). After all, the quadratic formula gives as a solution \[x=\dfrac{-b\pm\sqrt{D}}{2a}\] with \[D=b^2-4ac\tiny.\] In this problem with the equation \[3\, x^2-4\, x+1=0\] we have: \[a=3,\quad b=-4,\quad c=1,\quad D=(-4)^2 -4\times 3\times 1= 4\tiny.\] Because \(D\gt 0\), the number of solutions of the equation \(3\, x^2-4\, x+1=0\) is equal to \(2\).
The answer is according to the quadratic formula (possibly, after simplification): \[ x={{1}\over{3}}\quad\vee\quad x=1\tiny.\]
The quadratic formula
