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One of the roots is given. Find the other root.$6 x^{2}-x+c=0 ;-\frac{2}{3}$

$\frac{5}{6}$

Algebra

Chapter 5

QUADRATIC FUNCTIONS AND COMPLEX NUMBERS

Section 7

Sum and Product of the Roots of a Quadratic Equation

Equations and Inequalities

Quadratic Functions

Complex Numbers

Polynomials

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

Idaho State University

Lectures

01:32

In mathematics, the absolu…

01:11

01:07

One of the roots is given.…

00:57

00:47

01:04

00:58

01:10

04:12

For what value of $c$ will…

00:55

Fill in the blanks.If …

02:13

Use the discriminant to de…

02:45

One root of the polynomial…

01:58

00:43

Determine how many roots e…

02:06

Discuss the possibilities …

02:40

Use the fact that $x-2$ is…

02:31

The equation $x^{2}-3 x+a=…

01:56

One root of the equation $…

02:49

01:39

Find the roots of $x(x+6)(…

02:24

Use the Rational Root Theo…

have this quadratic equation. We don't have the C value but were given that one of the routes is equal to negative 2/3. We're going to use the fact that if we add the two routes, we get negative. Be over a. So just say negative 2/3 is a route we know that we don't was caught route to is equal to negative B. Well, negative negative one, because there's like there's a negative one here. Negative, negative one is a positive one, and a value is six. Then we're gonna use this to find the other route. We got to get rid of a negative 2/3 by adding 2/3. It's just write it like this. It's a little bit easier to both sides of the equation, except 2/3 if I most. Why numerator and denominator right to that is equal to 46 which is much more helpful because 16 plus four six is equal to 56 and that is equal to our route that we did not know 56

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