# Week 6 – precalculus 11

This week in Precalculus 11, we learned how to develop and use the Quadratic Formula. In our learning, quadratic equations take the form of, $ax^2 + bx + c = 0$. If we complete the square of this equation, we get another formula that can be used to determine solutions to any and all quadratic equations written in the same form as the original equation before we completed the square. This is called the quadratic formula. It looks like, $x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$ you can now plug in the numbers of the quadratic equation into this formula to find X. This eliminated the need to factor or use any other method. I will now show you how to solve a quadratic equation using this method.

First we must analyze the original equation. $x^2 + 4x - 3 = 0$

We can compare this with the basic function of, $ax^2 + bx + c = 0$

In this case, a is 1, b is 4, and c is -3. We can now plug these numbers into the quadratic formula. $x = \frac {-b \pm \sqrt {b^2 - 4ac}}{2a}$

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\/ $x = \frac {-(4) \pm \sqrt {(4)^2 - 4(1)(-3)}}{2(1)}$

Now we simplify $x = \frac {-4 \pm \sqrt {16 + 12}}{2}$

Then $x = \frac {-4 \pm \sqrt {28}}{2}$

and, $x = \frac {-4 \pm 2\sqrt {7}}{2}$

Take away the denominator $x = {-2 \pm \sqrt {7}}$

This would be the final answer for x. If the radicand were negative it would be extraneous.

This would make the roots: $x = {-2 + \sqrt {7}}$

and $x = {-2 - \sqrt {7}}$

If this were a graph, x would be intercepted twice.