Before you get started, take this readiness quiz.
When we solved quadratic equations in the last section by completing the square, we took the same steps every time. By the end of the exercise set, you may have been wondering ‘isn’t there an easier way to do this?’ The answer is ‘yes’. Mathematicians look for patterns when they do things over and over in order to make their work easier. In this section we will derive and use a formula to find the solution of a quadratic equation.
We have already seen how to solve a formula for a specific variable ‘in general’, so that we would do the algebraic steps only once, and then use the new formula to find the value of the specific variable. Now we will go through the steps of completing the square using the general form of a quadratic equation to solve a quadratic equation for \(x\).
We start with the standard form of a quadratic equation and solve it for \(x\) by completing the square.
| \(ax^2 + bx + c = 0, \quad a \ne 0\) | |
| Isolate the variable terms on one side. | \(ax^2 + bx \quad = -c\) |
| Make the coefficient of \(x^\) equal to \(1\), by dividing by \(a\). | \(\dfrac + \dfracx \quad = -\dfrac\) |
| Simplify. | \(x^2+ \dfracx \quad = -\dfrac\) |
| To complete the square, find \(\left(\dfrac \cdot \dfrac\right)^\) and add it to both sides of the equation. | |
| \(\left(\dfrac \dfrac\right)^=\dfrac | \(x^2 + \dfracx +>>< = -\dfrac\,+\,>>>>\) |
| The left side is a perfect square, factor it. | \(\left( x + \dfrac\right)^2 = -\dfrac+\dfrac\) |
| Find the common denominator of the right side and write equivalent fractions with the common denominator. | \(\left( x + \dfrac\right)^2 =\dfrac-\dfrac>>\) |
| Simplify. | \(\left( x + \dfrac\right)^2 =\dfrac-\dfrac\) |
| Combine to one fraction. | \(\left( x + \dfrac\right)^2 =\dfrac\) |
| Use the square root property. | \(x + \dfrac= \pm\sqrt<\dfrac>\) |
| Simplify the radical. | \(x + \dfrac= \pm\dfrac<\sqrt>\) |
| Add \(-\dfrac\) to both sides of the equation. | \(x = -\dfrac \pm\dfrac<\sqrt>\) |
| Combine the terms on the right side. | \(x = \dfrac<-b\pm\sqrt>\) |
The final equation is called the "Quadratic Formula."
The solutions to a quadratic equation of the form \(a x^+b x+c=0\), where \(a≠0\) are given by the formula:
To use the Quadratic Formula, we substitute the values of \(a,b\), and \(c\) from the standard form into the expression on the right side of the formula. Then we simplify the expression. The result is the pair of solutions to the quadratic equation.
Notice the Quadratic Formula (Equation \ref) is an equation. Make sure you use both sides of the equation.
Solve by using the Quadratic Formula: \(2 x^+9 x-5=0\).
Solution:
