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Problem 1 Problem 2 Problem 3 Problem 4 Problem 5 Problem 6 Problem 7 Problem 8 Problem 9 Problem 10 Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 Problem 16 Problem 17 Problem 18 Problem 19 Problem 20 Problem 21 Problem 22 Problem 23 Problem 24 Problem 25 Problem 26 Problem 27 Problem 28 Problem 29 Problem 30 Problem 31 Problem 32 Problem 33 Problem 34 Problem 35 Problem 36 Problem 37 Problem 38 Problem 39 Problem 40 Problem 41 Problem 42 Problem 43 Problem 44 Problem 45 Problem 46 Problem 47 Problem 48

Problem 41 Hard Difficulty

Write a quadratic equation with integer coefficients for each pair of roots.
$3+i, 3-i$

Answer

$x^{2}-6 x+10=0$

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Algebra

Algebra 2 and Trigonometry

Chapter 5

QUADRATIC FUNCTIONS AND COMPLEX NUMBERS

Section 7

Sum and Product of the Roots of a Quadratic Equation

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Watch More Solved Questions in Chapter 5

Problem 1
Problem 2
Problem 3
Problem 4
Problem 5
Problem 6
Problem 7
Problem 8
Problem 9
Problem 10
Problem 11
Problem 12
Problem 13
Problem 14
Problem 15
Problem 16
Problem 17
Problem 18
Problem 19
Problem 20
Problem 21
Problem 22
Problem 23
Problem 24
Problem 25
Problem 26
Problem 27
Problem 28
Problem 29
Problem 30
Problem 31
Problem 32
Problem 33
Problem 34
Problem 35
Problem 36
Problem 37
Problem 38
Problem 39
Problem 40
Problem 41
Problem 42
Problem 43
Problem 44
Problem 45
Problem 46
Problem 47
Problem 48

Video Transcript

we have a root equal to three plus I and another one You put a three minus I. And so this is actually, although it might look complex of lots gonna cancel out. For instance, when I add these 23 plus I and three minus I, you can see the eyes were gonna cancel out right away. And this is equal to negative. Be over a which whenever you can make a equal one, life will be easier. And so, like I said, we have a positive I in a negative I that cancel out three plus three is equal to six, and so six is equal to negative B if I must, by both sides of the negative one. B is equal to negative six. Now over here, let's find a product. Let's do three plus I and multiply it by three minus. I equal to see over a and remember is equal to one. So three times three is nine we got I times three as a positive three i in a negative items story and negative. Three of those will cancel and then we have I times a negative I Well, im times eyes negative one right, Remember, that's what we know about I squared equals negative one and so a negative I squared is a positive one equal to see over one. Well, that's see. So nine plus one is 10. There's RC Die. We can write the cigarette quadratic equation using a is one So x squared B is negative six So neg 06 x and a C of 10 senate equals zero rather.

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