Question

Solve by completing the square. z^2=8z-16

Answer 1

k so what I got is, z^2-8z=-16

then z^2-8z+16=0

Answer 2

but then, (z-4)^2=0

Answer 3

but I can't find the square root of 0 right?

Answer 4

Yes you are right..

Answer 5

You are on the right track. Find the square root of both sides.

Answer 6

You can.

Answer 7

\(\sqrt0 = 0\)

Answer 8

oh. z-4=0?!

Answer 9

Yes..

Answer 10

Exactly :)

Answer 11

so z = 4 and -4 right.

Answer 12

Oops.

Answer 13

No..

Answer 14

Add 4 both the sides..

Answer 15

only one answer?

Answer 16

You don't need to complicate it.

\( \color{Black}{\Rightarrow z - 4 + 4 = 0 + 4}\)

Answer 17

Yes. We have only one root in a perfect square :)

Answer 18

No you will get two but same answer..

Answer 19

oh and also, what happens when the right side of the equation is a negative? how will i do the square root then?

Answer 20

You may say that 4 is a root with multiplicity 2.

Answer 21

But there are no other distinct roots.

Answer 22

another question, I have (x+2)^2=-9...

Answer 23

Complex numbers kick in:

\( \color{Black}{\Rightarrow x^2 = -25 \Longrightarrow x = 5i }\)

Answer 24

\( \color{Black}{\Rightarrow x + 2 = 3i }\)

Answer 25

See basically we write them as:

\[(z-4)^2 = 0\]

\[(z-4)(z-4) = 0\]

\[z - 4 = 0 \qquad \And \qquad z - 4 = 0\]

z = 4 and z = 4..

Answer 26

You may find the square root of -9:

\( \color{Black}{\Rightarrow \sqrt{-9} = \sqrt{-1} \times \sqrt{9} = i \times 3 = 3i}\)

Answer 27

@waterineyes No need to complicate it there too. Just find the square root of both sides :)

Answer 28

Back to the point: \(x + 2 = 3i\) then you subtract 2 from both sides.

Answer 29

That is not applicable and not the correct method though @ParthKohli

Answer 30

Are you getting what parth has done @Steeben ??

Answer 31

@ you know what @ParthKohli has done it totally wrong..

He converted the quadratic into linear and only found one solution but the equation has two solutions..

Answer 32

No!

Answer 33

The equation has only one root with multiplicity 2.

Answer 34

I can prove my point..

Answer 35

okay guys... no need to fight, ... BE HAPPY LOL

k so I started with this:

Answer 36

You may find the square root in case of a perfect square.

Answer 37

No..
this is complex one dear and not real..

Answer 38

x^2+4x+13=0

Answer 39

x^2+4x=-13
x^2+4x+4=-13+4

Answer 40

Yes you are on a right track @Steeben

Answer 41

(x+2)^2=-9

Answer 42

No this one is not right..

Answer 43

then then i dont' know what to do... supposedly 9 is going to be 3 but it's negative :/

Answer 44

I already showed how \(\sqrt{-9} = 3i\)

Answer 45

See,

\(x^2 + 4x + 13 = 0\)

You got this one??

Answer 46

yes

Answer 47

@ParthKohli how did you get a random variable i?

Answer 48

@ParthKohli please let me explain my point..

Answer 49

\( \color{Black}{\Rightarrow x^2 + 4x = -13 }\)

\( \color{Black}{\Rightarrow x^2 + 4x + 4 = -13 + 4}\)

\( \color{Black}{\Rightarrow (x + 2)^2 = -9}\)

That is the complex number \(i \).

Answer 50

\(i = \sqrt{-1}\)

Answer 51

@Steeben I am telling you the correct way to do this..

Can you listen to me or not??

Answer 52

okay just type it out

Answer 53

Using quadratic formula:

\(D = b^2 - 4ac\)

D = 16 - 52 = -36

\[x = \frac{-4 \pm \sqrt{-36i}}{2}\]

See, \(\sqrt{-36} = \sqrt{(-1) \times 36} = 6i\)

Answer 54

Put in the equation and solve:

\[x = -2 \pm 3i\]

So this has two solutions:

\(x = -2 + 3i\) and \(x = -2 - 3i\)

Answer 55

@ParthKohli God has given you eyeballs check with them and tell me are they the same root with multiplicity of 2???

Answer 56

Okay. You win, I lose.

Answer 57

That is better..
Ha ha ha..

Answer 58

It is not about win and loss it is about knowledge..

Answer 59

uhh i'm kinda stupid so i don't know anything XD.. but uhm i don't even know what multiplicity of 2 means same root whatever thing.

but uh i still don't get the i.. I know the quad formula but the i I just don't know where that came from or what it's for.

Answer 60

You want to know about i??

Answer 61

my teacher says that if there's a negative square root it's not solvable.

Answer 62

Yes it is the case when we have only real numbers in our Mathematics..

Then I don't know who came he added one more chapter to our Mathematics and that chapter is called Complex Numbers..

See, value of i is given as:

\[i = \sqrt{-1}\]

Can you remember this??

Basically the concept of i is brought to solve for negative values in the square roots..

\(i\) is basically called \(iota\)..

Getting or not??

Answer 63

See, \(\sqrt{36} = 6\)

\(\sqrt{-36} = 6i\)

Answer 64

so i is kinda "replacing" the negative?

Answer 65

The logic or trick is :

Whenever there are negative numbers inside the square root then take the square root of the number and multiply the number with i..


Example: \(\sqrt{-114}\)

Can you tell me the square root of 144?? @Steeben

Answer 66

12i

Answer 67

Yes nice..
You are learning very fast..

Answer 68

I show you why \(\sqrt{-144} = 12i\)

See first you have to remember that"

\[\sqrt{ab} = \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\]

Answer 69

so (x+2)^2=-9 and then x+2 = 3i

Answer 70

You cannot solve it directly, the correct way I have done above using Quadratic Formula..