## 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 $x^2 - 2x - 3 = 0$?

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

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{aligned} x &= \frac{-(-4) \pm \sqrt{(-4)^2-4(1)(5)}}{2(1)} \\ x &= \frac{4 \pm \sqrt{-4}}{2} \\ x &= 2 \pm i \end{aligned}

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 unless the discriminant is equal to 0. So,

\begin{aligned} b^2 - 4ac &= 0\\ (7)^2 - 4(2)c &= 0\\ c &= \tfrac{49}{8} \end{aligned} _\square

### $x$ is a negative number 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{aligned} 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{aligned}

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

This produces the following quadratic equation:

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

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

Note by Arron Kau
6 years, 8 months ago

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