Lynn Marecek; MaryAnne Anthony-Smith; Andrea Honeycutt Mathis; Jay Abramson; Sharon North; Amy Baldwin; Alyssa Howell

For 1 - 4, use the following information:

A parabola crosses the $x$ -axis at (-1,0) and (2,0). We want to write a quadratic equation with those $x$ -intercepts. We can do this without graphing.

1.

When $y = 0$ , what must $x$ be?

Compare your answer:

We know that (-1,0) and (2,0) are points on the graph when $y = 0$ . So there are two possibilities for $x$ : $x = - 1$ and $x = 2$ .

$x = - 1$ and $x = 2$ are called the real solutions of the quadratic because they are the solutions of the equation when we set $y = 0$ .

2.

Rewrite the equations from number 1 so they both equal zero.

Compare your answer:

$x + 1 = 0$ and $x - 2 = 0$ .

Because each of the equations equals zero, multiplying them together will also equal zero.

3.

What equation do you get when you multiply the previous equations together?

Compare your answer:

$(x + 1) (x - 2) = 0$

4.

Multiply the binomials.

Compare your answer:

$x^{2} - x - 2 = 0$

This answer can also give us the more general quadratic with those $x$ -intercepts. We replace 0 with $y$ to show that $y$ varies and isn’t always 0. So, $y = x^{2} - x - 2$ is a parabola that crosses the $x$ -axis at -1 and 2.

Self Check

The solution to a quadratic equation is $x = 4$ and $x = - 8$ . What is the equation?

Additional Resources

When you write a quadratic in factored form, that helps you find the $x$ -intercepts.

$y = x^{2} + 2 x - 3$

$y = (x + 3) (x - 1)$

When we set $y = 0$ , we find the places where the parabola crosses the $x$ -axis.

$0 = (x + 3) (x - 1)$

So, $0 = x + 3$ and $0 = x - 1$

$x = - 3$ and 1.

-3 and 1 are called the real solutions to the quadratic equation.

We can also start with the real solutions and use them to write quadratic equations.

When we know that $x = - 3$ and 1, then that means $0 = x + 3$ and $0 = x - 1$ .

So, $0 = (x + 3) (x - 1)$ .

And, using the FOIL method to multiply this out, we come to

$0 = x^{2} + 2 x - 3 .$

When $y$ is not equal to zero, we have the following quadratic equation

$y = x^{2} + 2 x - 3$ , which is exactly where we started!

Try it

Try It: Writing Quadratic Equations from Real Solutions

1.

Write a quadratic from the real solutions $x = 1$ and $x = 7$ without graphing.

Compare your answer: $x^{2} - 8 x + 7$

$x = 1$ and $x = 7$ become $x - 1 = 0$ and $x - 7 = 0$ . $y = (x - 1) (x - 7) = x^{2} - x - 7 x + 7 = x^{2} - 8 x + 7$ .

7.13.4 Writing Quadratic Equations from Real Solutions

Additional Resources

Try It: Writing Quadratic Equations from Real Solutions