## Definition

The **quadratic formula** gives us the solutions, or roots, to a quadratic equation of the form \(ax^2 + bx + c\):

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

## Technique

## What is the sum of the possible solutions of \( x^2 - 2x - 3 = 0 \)?

Using the quadratic formula:

\[ \begin{align} x &= \frac{-(-2) \pm \sqrt{(2)^2 - 4(1)(-3)}}{2(1)} \\ x &= 1 \pm 2 \\ x &=3 \text{ or } x=-1 \end{align} \]

The the answer is \( 3-1=2 \). \( _\square \)

The quadratic formula is helpful even when the solutions are complex numbers:

## The roots of \( x^2 - 4x + 5 \) are two complex numbers, \( z_1 \) and \( z_2 \). What is the sum of \( z_1 \) and \( z_2 \)?

\[ \begin{align} x &= \frac{-(-4) \pm \sqrt{(-4)^2-4(1)(5)}}{2(1)} \\

x &= \frac{4 \pm \sqrt{-4}}{2} \\ x &= 2 \pm i \end{align} \]Thus, the two roots are \( z_1 = 2-i \) and \( z_2 = 2 + i \) and their sum is \( (2-i) + (2 + i ) = 4 \). \( _\square \)

Whether the roots are real or complex depends on the quadratic formula's **discriminant**, \( b^2 - 4ac\), the expression inside the square root. The roots are real when the discriminant is positive and complex when the discriminant is negative.

## Application and Extensions

## For what value of \(c\) will \( 2x^2 + 7x + c = 0\) have only a single real root?

The quadratic formula's \(\pm\) tells us that there will always be two roots

unlessthe discriminant is equal to 0. So,\[ \begin{align} b^2 - 4ac &= 0\\

(7)^2 - 4(2)c &= 0\\ c &= \tfrac{49}{8} \end{align} _\square \]

## \(x\) is a

negativenumber such that \( x^2+9x -22 = 0 \). What is the sum of all possible values of \(y\) which satisfy the equation \( x = y^2 - 13y + 24 \)?Since \( x^2+9x -22 = 0 \), we know that

\[ \begin{align} x &= \frac{-9 \pm \sqrt{81-4(-22)}}{2} \\ &= \frac{-9 \pm \sqrt{169}}{2} \\ &= \frac{-9 \pm 13}{2} \\ &= 2 \text{ or } -11 \end{align} \]

Since \( x \) is negative, \( x=-11 \). So now we need to solve \( -11 = y^2 - 13y + 24 \).

This produces the following quadratic equation:

\[ \begin{align} y^2 - 13y + 35 &=0 \\ (y-5)(y-7)&=0 \end{align} \]

Thus \( y=5 \text{ or } 7\), and the answer is 12. \( _\square \)

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