Day 149 – April 26

Algebra: Chapter 13, Lesson 4, page 589.

The Quadratic Formula, finally!

Given that ${\displaystyle a{x}^2+bx+c=0}$, then the quadratic formula

${\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}}$

gives the solutions of the quadratic equation. This requires that the quadratic equation is always in standard form:

• ${\displaystyle a}$ is the coefficient of the ${\displaystyle x^2}$ term
• ${\displaystyle b}$ is the coefficient of the ${\displaystyle x}$ term and
• ${\displaystyle c}$ is the constant

Memorize it and memorize the discriminant, the expression under the radical (the ${\displaystyle b^2-4ac}$ thingy).

• If the discriminant is > 0, then there are 2 real number solutions
• If the discriminant is = 0, then there is just 1 real number solution
• If the discriminant is < 0, then there are NO real number solutions because you don’t know (yet) how to take the root of a negative real number.

Don’t forget that the

• solutions
• answers
• x-intercepts
• roots and
• zeroes

all mean the same thing. By definition, the equation ${\displaystyle a{x}^2+bx+c=0}$ implies that we are setting ${\displaystyle y=0}$ and finding the x-intercepts or the roots or the answers or the solutions!!

Once again, Purplemath.com comes to the rescue, check out these examples for the use of the quadratic formula!

Two of tonight’s homework problems solved by MrE are here! Just click it.

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Algebra 1a: CST Review

TEST PRACTICE #1 – today in class!