7 y^2-2 y+3=0 | Numerade

Name: Problem 52, Chapter 8: Elementary and Intermediate Algebra
Uploaded: 2021-09-24T00:50:08-08:00
Duration: 2 min 31 s
Channel: Julie Silva
Description: 7 y^2-2 y+3=0

Key Concepts

Complex Numbers

Complex numbers extend the real numbers and are used to represent the roots of quadratic equations when the discriminant is negative. In these cases, the square root of the negative number introduces the imaginary unit 'i', leading to solutions in the form of complex conjugates. Complex numbers are an essential part of algebra that allow for the complete solution of polynomial equations and have many applications in various fields of science and engineering.

Quadratic Formula

The quadratic formula is a universal method for solving any quadratic equation. It is derived by completing the square and provides the roots as (-b ± ?(b² - 4ac)) / (2a). This formula is essential because it allows for the direct computation of the roots regardless of whether the equation can be factored easily.

Quadratic Equations

Quadratic equations are polynomial equations of degree two, typically written in the standard form ax² + bx + c = 0. They play a fundamental role in algebra and appear in various areas of mathematics, physics, and engineering. The solutions to quadratic equations are known as the roots of the equation and can be real or complex numbers.

Discriminant

The discriminant of a quadratic equation, given by b² - 4ac, is a key component that indicates the nature of the roots. A positive discriminant results in two distinct real roots, zero yields a repeated real root, and a negative discriminant leads to two complex conjugate roots. Understanding the discriminant is crucial for predicting the behavior of the solutions without fully solving the equation.

Transcript

00:01 So in part a of this problem, we're being asked to factor the given trinomial.

00:04 Well, in factoring, you should always start by looking for a greatest common factor, but in this case, there isn't one.

00:09 So i see we have a trinomial here.

00:11 So we're going to have to factor by trial and error.

00:14 So i'm going to set up our two binomials here.

00:18 Now i noticed that our first coefficient is positive 20.

00:21 Well, what are our factors of 20? one in 20, 2 in 10, and 4 and 5.

00:27 And because it's positive, we're going to keep all of those positive.

00:30 Now, we also need to figure out the factors of our constant negative 3.

00:34 Well, that would either be 1 in negative 3 or negative 3 and 1.

00:38 But because 3 is prime, we know that they're going to end with a 3 in the 1.

00:42 So we'll at least start with that there.

00:44 So now, what we have to do is to come up with some combination, so that way our inside or our middle terms will equal to 7.

00:52 Well, in this case, i'm going to look at the numbers 4 and 5.

00:55 Because what i know is that if i have 4y here, i'd have 3 times 4y, which is 12y.

01:01 And if i have 5 y at the beginning, for my outside terms, i'd have 5 times y, which is 5 y.

01:07 And using these two terms, i can get a positive 7 if the 12 is positive and the 5 is negative.

01:13 So now let's just put in our signs.

01:16 Well, the 3 times 4, that needs to be positive, so we'll put a plus sign here.

01:20 And 5 times 1 need to be negative...

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