Quadratic Zeros

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quadratic formula

Our library has two types of Elementary Functions, so far:

Linear functions
Quadratic functions

We have also discovered one of our goals for analysis is to identify zeros of functions.

Unless the linear function is actually a constant function (which is a linear function), linear functions have exactly one zero. We can identify this unique zero by setting the formula equal to $0$ and solving. This is accomplished by combining like terms and isolating the variable on one side of the equation.

We have seen one approach to identifying zeros of quadratic functions.

Completing the Square

We have a second approach.

The Quadratic Formula

After stepping through the procedure of completing the square, we could have extended the procedure and solved for $t$.

Completing the square gave us

\[ a \, t^2 + b \, t + c = a \left (t + \frac {b}{2 a} \right )^2 + c - \frac {b^2}{4 a} \]

Now, pretend this was equal to $0$ in a quadratic equation.

\[ a \, t^2 + b \, t + c = 0 \]

\[ a\left (t + \frac {b}{2 a}\right )^2 + c - \frac {b^2}{4 a} = 0\]

Solve for $t$.

\[ a \left (t + \frac {b}{2 a} \right )^2 = \frac {b^2}{4 a} - c\]

\[ \left (t + \frac {b}{2 a} \right )^2 = \frac {b^2}{4 a^2} - \frac {c}{a}\]

\[ \left (t + \frac {b}{2 a} \right )^2 = \frac {\answer {b^2 - 4 a c}}{4 a^2} \]

Either

\[ t + \frac {b}{2 a} = \sqrt {\frac {b^2 - 4 a c}{4 a^2}} = \frac {\sqrt {b^2 - 4 a c}}{| 2a |} \]

\[ t + \frac {b}{2 a} = -\sqrt {\frac {b^2 - 4 a c}{4 a^2}} = -\frac {\sqrt {b^2 - 4 a c}}{| 2a |} \]

If $a > 0$, then $| 2a | = 2a$.
If $a < 0$, then $| 2a | = -2a$.

Either way, we still get one fraction using a negative sign and one fraction fraction not using a negative sign. Therefore, we can drop the absolute value signs.

Either

\[ t + \frac {b}{2 a} = \sqrt {\frac {b^2 - 4 a c}{4 a^2}} = \frac {\sqrt {b^2 - 4 a c}}{2a} \]

\[ t + \frac {b}{2 a} = -\sqrt {\frac {b^2 - 4 a c}{4 a^2}} = -\frac {\sqrt {b^2 - 4 a c}}{2a} \]

And, finally...

Either

\[ t = - \frac {b}{2 a} + \frac {\sqrt {b^2 - 4 a c}}{2a} = \frac {-b + \sqrt {b^2 - 4 a c}}{2a} \]

\[ t = - \frac {b}{2 a} -\frac {\sqrt {b^2 - 4 a c}}{2a} = \frac {-b - \sqrt {b^2 - 4 a c}}{2a} \]

People generally shorthand these two separate solutions as

\[ t = \frac {-b \pm \sqrt {b^2 - 4 a c}}{2a} \]

This is known as The Quadratic Formula.

The Quadratic Formula

If $a \, t^2 + b \, t + c = 0$ with $a \ne 0$, then

\[ t = - \frac {b}{2 a} + \frac {\sqrt {b^2 - 4 a c}}{2a} = \frac {-b + \sqrt {b^2 - 4 a c}}{2a} \]

\[ t = - \frac {b}{2 a} -\frac {\sqrt {b^2 - 4 a c}}{2a} = \frac {-b - \sqrt {b^2 - 4 a c}}{2a} \]

Shorthand:

\[ t = \frac {-b \pm \sqrt {b^2 - 4 a c}}{2a} \]

Let’s apply this to the previous three examples

Two Real Solutions

Solve $4 t^2 - 4 t - 8 = 0$

explanation

To use the Quadratic Formula the quadratic equation has to have everything on one side and $0$ on the other. Our equation is already in that form.

Apply the Quadratic Formula

To use the Quadratic Formula, we match our quadratic to the general quadratic template:

\[ 4 t^2 - 4 t - 8 = 0 = a \, t^2 + b \, t + c \]

Matching the leading terms tells us that $a = 4$.
Matching the linear terms tells us that $b = -4$.
Matching the constant terms tells us that $c = -8$.

These are the substitutions we need to perform in the Quadratic Formula.

\[ t = \frac {-(-4) \pm \sqrt {(-4)^2 - 4 (4) (-8)}}{2 (4)} \]

\[ t = \frac {4 \pm \sqrt {16 + 128}}{8} \]

\[ t = \frac {4 \pm \sqrt {\answer {144}}}{8} \]

\[ t = \frac {4 \pm \answer {12}}{8} \]

Either $t = \frac {4 + 12}{8} $ or $t = \frac {4 - 12}{8} $

Either $t = \frac {16}{8} $ or $t = \frac {-8}{8} $

Either $t = \answer {2}$ or $t = \answer {-1}$

One Real Solution

Solve $2 x^2 - 12x + 21 = 3$

explanation

First, get everything to one side and $0$ on the other side.

\[ 2 x^2 - 12x + 18 = 0 \]

Apply the Quadratic Formula

\[ t = \frac {-(-12) \pm \sqrt {(-12)^2 - 4 (2) (18)}}{2 (2)} \]

\[ t = \frac {12 \pm \sqrt {144 - 144}}{4} \]

\[ t = \frac {12 \pm \answer {0}}{4} \]

If you add or subtract $0$, you get the same result.

\[ t = \frac {12}{4} = 3 \]

No Real Solutions

Solve $2 m^2 - 12m + 21 = 1$

explanation

First, get everything to one side and $0$ on the other side.

\[ 2 m^2 - 12m + 20 = 0 \]

Apply the Quadratic Formula

\[ m = \frac {-(-12) \pm \sqrt {(-12)^2 - 4 (2) (20)}}{2 (2)} \]

\[ m = \frac {12 \pm \sqrt {144 - 160}}{4} \]

\[ m = \frac {12 \pm \sqrt {\answer {-16}}}{4} \]

The real numbers don’t hold a number equal to $\sqrt {-16}$. Therefore, there are no real solutions.

In each of the examples above the number of solutions was determined by $\sqrt {b^2 - 4 a c}$. The inside of the square root, $b^2 - 4 a c$, is called the discriminant and its sign tells us how many real solutions the equation has.

If $b^2 - 4 a c$ $>$$=$$<$ $0$, then there are two distinct real solutions.
If $b^2 - 4 a c$ $>$$=$$<$ $0$, then there is one real solutions.
If $b^2 - 4 a c$ $>$$=$$<$ $0$, then there are no distinct real solutions.

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more examples can be found by following this link
More Examples of Quadratics

2026-05-25 15:13:44