Investigation: Deriving the quadratic formula
Recall that some quadratic equations can be solved by factoring or by completing the square. But how do we solve a quadratic equation that cannot be factored? Is there a shortcut that we could use so that we don't have to perform the process of completing the square each time?
Exploration
Let's consider the following quadratic equations:
Equation 1: x^{2} + 5 x + 6 = 0
Equation 2: 3 x^{2} + 5 x - 11 = 0
- Solve each of the quadratic equations by completing the square.
- Which of the above equations would be best solved using a different method?
- What patterns do you notice when solving for a quadratic by completing the square?
- How might we represent the solution to any quadratic equation?
To consider the solution to any quadratic equation, let's first consider how we might represent quadratic equations generally. That is, with the equation a x^{2} + bx + c = 0, where a \neq 0.
We say that a \neq 0 because, if it were, we wouldn't have an x^{2} term and the equation wouldn't be quadratic. However, the parameter a can be any other real number, and b or c can be any real number without restriction.
Now, let's consider how we might find the solution to any quadratic equation. Try answering the question below, then reveal the solution to see if you were on the right track.
Consider the quadratic equation $ax^2+bx+c=0$ax2+bx+c=0, where $a\ne0$a≠0.
Solve for $x$x by completing the square.
Consider the quadratic equation $ax^2+bx+c=0$ax2+bx+c=0, where $a\ne0$a≠0.
Solve for $x$x by completing the square.
The formula in the solution above can be simplified to a formula known as the Quadratic Formula.
Given a quadratic equation a x^{2} + bx + c = 0, where a, b, and c are real numbers and a \neq 0,
the solutions to the equation can be found using the quadratic formula,
x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}
There are many different ways to derive the quadratic formula. Another derivation is in the problem below. After reading the proof for understanding, try and find a reason for each step of the proof.
Consider the following steps for a proof of the Quadratic Formula.
$ax^2+bx+c$ax2+bx+c
$=$=
$0$0
$4a^2x^2+4abx+4ac$4a2x2+4abx+4ac
$=$=
$0$0
$4a^2x^2+4abx+4ac+b^2$4a2x2+4abx+4ac+b2
$=$=
$b^2$b2
$4a^2x^2+4abx+b^2$4a2x2+4abx+b2
$=$=
$b^2-4ac$b2−4ac
$\left(2ax+b\right)^2$(2ax+b)2
$=$=
$b^2-4ac$b2−4ac
$2ax+b$2ax+b
$=$=
$\pm\sqrt{b^2-4ac}$±√b2−4ac
$2ax$2ax
$=$=
$-b\pm\sqrt{b^2-4ac}$−b±√b2−4ac
$x$x
$=$=
$\frac{-b\pm\sqrt{b^2-4ac}}{2a}$−b±√b2−4ac2a
Explain what was done to produce each step.
Consider the following steps for a proof of the Quadratic Formula.
| $ax^2+bx+c$ax2+bx+c | $=$= | $0$0 |
| $4a^2x^2+4abx+4ac$4a2x2+4abx+4ac | $=$= | $0$0 |
| $4a^2x^2+4abx+4ac+b^2$4a2x2+4abx+4ac+b2 | $=$= | $b^2$b2 |
| $4a^2x^2+4abx+b^2$4a2x2+4abx+b2 | $=$= | $b^2-4ac$b2−4ac |
| $\left(2ax+b\right)^2$(2ax+b)2 | $=$= | $b^2-4ac$b2−4ac |
| $2ax+b$2ax+b | $=$= | $\pm\sqrt{b^2-4ac}$±√b2−4ac |
| $2ax$2ax | $=$= | $-b\pm\sqrt{b^2-4ac}$−b±√b2−4ac |
| $x$x | $=$= | $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$−b±√b2−4ac2a |
Explain what was done to produce each step.
Extension questions
- Do some research to find out when and how the quadratic formula was originally established.
- Think of some other ways to derive the quadratic formula.
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Are there formulas for cubics, quartics or higher degree polynomials? What is the greatest degree polynomial that can be solved in general using a formula?