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Quadratic Formula

Prerequisites

Definition of a Quadratic Polynomial | \(f(x) = a \cdot x^{2} + b \cdot x + c\)

Description

Can be used to find the roots of a Quadratic Polynomial.

\[\htmlClass{sdt-0000000003}{x}_{1, 2} = \frac{-\htmlClass{sdt-0000000033}{b} \pm \sqrt{\htmlClass{sdt-0000000033}{b}^{2} - 4 \htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000004}{c}}}{2 \htmlClass{sdt-0000000121}{a}}\]

Symbols Used:

This is a symbol for any generic variable. It can hold any value, whether that be an integer or a real number, or a complex number, or a matrix etc.
\( c \)

This is a symbol for any tertiary generic constant. It can hold any numerical value
\( b \)

This is a symbol for any secondary generic constant. It can hold any numerical value
\( a \)

This is a symbol for any generic constant. It can hold any numerical value

Derivation

  1. Consider the definition of a quadratic polynomial:
    \[\htmlClass{sdt-0000000120}{f(x)} = \htmlClass{sdt-0000000121}{a} \cdot \htmlClass{sdt-0000000003}{x}^{2} + \htmlClass{sdt-0000000033}{b} \cdot \htmlClass{sdt-0000000003}{x} + \htmlClass{sdt-0000000004}{c}\].
    We will set the value to zero, so
    \[\htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000003}{x}^2 + \htmlClass{sdt-0000000033}{b} \htmlClass{sdt-0000000003}{x} + \htmlClass{sdt-0000000004}{c} = 0\]
  2. From here, we will divide both sides by \(\htmlClass{sdt-0000000121}{a}\), assuming it is never equal to \(0\), yielding...
    \[\htmlClass{sdt-0000000003}{x}^2 + \frac{\htmlClass{sdt-0000000033}{b}}{\htmlClass{sdt-0000000121}{a}}\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000004}{c}}{\htmlClass{sdt-0000000121}{a}} = 0\]
  3. We can now subtract \(\frac{\htmlClass{sdt-0000000004}{c}}{\htmlClass{sdt-0000000121}{a}}\) from both sides. This gives us:
    \[\htmlClass{sdt-0000000003}{x}^2 + \frac{\htmlClass{sdt-0000000033}{b}}{\htmlClass{sdt-0000000121}{a}}\htmlClass{sdt-0000000003}{x} = -\frac{\htmlClass{sdt-0000000004}{c}}{\htmlClass{sdt-0000000121}{a}}\]
  4. We will now add the term \(\frac{\htmlClass{sdt-0000000033}{b}^2}{4\htmlClass{sdt-0000000121}{a}^2}\) to both sides. You will see why in the next few steps. For now just accept it, as we can add anything we like to both sides.
    \[\htmlClass{sdt-0000000003}{x}^2 + \frac{\htmlClass{sdt-0000000033}{b}}{\htmlClass{sdt-0000000121}{a}}\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000033}{b}^2}{4\htmlClass{sdt-0000000121}{a}^2} = \frac{\htmlClass{sdt-0000000033}{b}^2}{4\htmlClass{sdt-0000000121}{a}^2}-\frac{\htmlClass{sdt-0000000004}{c}}{\htmlClass{sdt-0000000121}{a}}\]
  5. Let us draw our attention to the left hand side, notice that it is equivalent to \((\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000033}{b}}{2 \htmlClass{sdt-0000000121}{a}})^2\) becasuse...
    \[(\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000033}{b}}{2 \htmlClass{sdt-0000000121}{a}})^2 = (\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000033}{b}}{2 \htmlClass{sdt-0000000121}{a}})(\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000033}{b}}{2 \htmlClass{sdt-0000000121}{a}}) = \htmlClass{sdt-0000000003}{x}^2 + \frac{\htmlClass{sdt-0000000033}{b}}{\htmlClass{sdt-0000000121}{a}}\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000033}{b}^2}{4\htmlClass{sdt-0000000121}{a}^2} \]
  6. We can now simplify the equation demonstrated in step (4) to get:
    \[(\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000033}{b}}{2\htmlClass{sdt-0000000121}{a}})^2 = \frac{\htmlClass{sdt-0000000033}{b}^2}{4\htmlClass{sdt-0000000121}{a}^2}-\frac{\htmlClass{sdt-0000000004}{c}}{\htmlClass{sdt-0000000121}{a}}\]
  7. We can now multiply the numerator and denominator of our \(\frac{\htmlClass{sdt-0000000004}{c}}{\htmlClass{sdt-0000000121}{a}}\) term by \(4 \htmlClass{sdt-0000000121}{a}\) to get:
    \[(\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000033}{b}}{2\htmlClass{sdt-0000000121}{a}})^2 = \frac{\htmlClass{sdt-0000000033}{b}^2}{4\htmlClass{sdt-0000000121}{a}^2}-\frac{4\htmlClass{sdt-0000000121}{a}\htmlClass{sdt-0000000004}{c}}{4\htmlClass{sdt-0000000121}{a}^2}\]
  8. We can now contract the right hand side into one expression, as they have the same denominator, this gives us:
    \[(\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000033}{b}}{2\htmlClass{sdt-0000000121}{a}})^2 = \frac{\htmlClass{sdt-0000000033}{b}^2 - 4\htmlClass{sdt-0000000121}{a}\htmlClass{sdt-0000000004}{c}}{4\htmlClass{sdt-0000000121}{a}^2}\]
  9. We can now take the square root of both sides, to get:
    \[\htmlClass{sdt-0000000003}{x} + \frac{\htmlClass{sdt-0000000033}{b}}{2\htmlClass{sdt-0000000121}{a}} = \frac{\pm\sqrt{\htmlClass{sdt-0000000033}{b}^2 - 4\htmlClass{sdt-0000000121}{a}\htmlClass{sdt-0000000004}{c}}}{\sqrt{4\htmlClass{sdt-0000000121}{a}^2}}\]
  10. We can now simplify the right hand side and move our \(\frac{\htmlClass{sdt-0000000033}{b}}{2\htmlClass{sdt-0000000121}{a}}\) to the other side to get...
    \[\htmlClass{sdt-0000000003}{x} = -\frac{\htmlClass{sdt-0000000033}{b}}{2\htmlClass{sdt-0000000121}{a}} + \frac{\pm\sqrt{\htmlClass{sdt-0000000033}{b}^2 - 4\htmlClass{sdt-0000000121}{a}\htmlClass{sdt-0000000004}{c}}}{2a}\]
  11. Finally, the right hand side can be contracted to give us:
    \[\htmlClass{sdt-0000000003}{x}_{1, 2} = \frac{-\htmlClass{sdt-0000000033}{b} \pm \sqrt{\htmlClass{sdt-0000000033}{b}^{2} - 4 \htmlClass{sdt-0000000121}{a} \htmlClass{sdt-0000000004}{c}}}{2 \htmlClass{sdt-0000000121}{a}}\]
    as required.

Example

Consider the quadratic equation: \(5\htmlClass{sdt-0000000003}{x}^2 - 10\htmlClass{sdt-0000000003}{x} -15 = 0\). What are the two solutions for \(\htmlClass{sdt-0000000003}{x}\)?

Values:

We can now use the quadratic formula to get:
\[\htmlClass{sdt-0000000003}{x}_{1, 2} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot (5) \cdot (-15)}}{2 \cdot 5}\]

This simplifies to:

\[\htmlClass{sdt-0000000003}{x}_{1, 2} = \frac{10 \pm \sqrt{400}}{10} = \frac{10 \pm 20}{10}\]

This means the 2 solutions are:

\[\htmlClass{sdt-0000000003}{x}_1 = \frac{30}{10} = 3, \:\:\: \text{and} \:\:\: \htmlClass{sdt-0000000003}{x}_2 = \frac{-10}{10} = -1\]