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One root of a quadratic equation is three more than the other. The sum of the roots is 15. Write the equation.

$x^{2}-15 x+\frac{675}{16}=0$

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

McMaster University

University of Michigan - Ann Arbor

Lectures

01:32

In mathematics, the absolu…

01:11

01:20

Write a quadratic equation…

00:20

Fill in the blanks.To …

01:06

01:04

02:03

Find two quadratic equatio…

00:45

03:14

Solve each equation by gra…

01:10

One of the roots is given.…

02:16

Solve the quadratic equati…

01:00

Without solving each equat…

02:04

01:34

00:29

01:22

02:13

An equation is given, foll…

02:18

02:32

Solve each equation by fin…

00:32

Explain how one could writ…

all right. Interesting. 60 situation here we have that one route is equal to three more than the other room. And so I'm just gonna say, OK, one Route X and the other one is X plus three to make it three more and you some algebra to kind of find that. And so we know also that the sun is equal to 15. So if I put all this in, I'm gonna get expose three plus X. I know it's gonna equal. Let me write that a little bit more nice. It's gonna equal 15. And so this basically becomes two. X Plus three is equal to 15 little bit of algebra here. Let's subtract three from both sides of the equation. Getting rid of that leaving us with two X is equal to 12. If I want to buy, let's do blue. If we multiply both sides or divide both sides by two were multiplied by half. However you prefer to think of it, we get an X value equal to six, meaning that one route is six plus three is nine and the other room is just six. And so knowing that we can play these in to get our see value. And so let's go ahead and do that by plugging in a six times and nine. And that's gonna equal. See over a And let's kind of talk about here because what are a value could be? Let's say, let's try to make a value one and see if this works. If a is one, then negative 15/1 equals This whole thing's RB value would be a negative 15 because negative 15/1. He put a negative there that would get that 15 that we want there. And so now let's use this. Nine times six is 54 is equal to see Asia one right? So that means 54 is our see value. And so let's go ahead and put this all together for a values one or equation is X squared B values negative. 15 0 excuse me. Negative 15 and let's go fishing to X and R C. Value is 54 and we're gonna go ahead and set that equal to zero and call it good

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