Ask your own question, for FREE!
Mathematics 35 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

k so what I got is, z^2-8z=-16 then z^2-8z+16=0

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Yes you are right..

Parth (parthkohli):

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

Parth (parthkohli):

You can.

Parth (parthkohli):

\(\sqrt0 = 0\)

OpenStudy (anonymous):

oh. z-4=0?!

OpenStudy (anonymous):

Yes..

Parth (parthkohli):

Exactly :)

OpenStudy (anonymous):

so z = 4 and -4 right.

Parth (parthkohli):

Oops.

OpenStudy (anonymous):

No..

OpenStudy (anonymous):

Add 4 both the sides..

OpenStudy (anonymous):

only one answer?

Parth (parthkohli):

You don't need to complicate it. \( \color{Black}{\Rightarrow z - 4 + 4 = 0 + 4}\)

Parth (parthkohli):

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

OpenStudy (anonymous):

No you will get two but same answer..

OpenStudy (anonymous):

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

Parth (parthkohli):

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

Parth (parthkohli):

But there are no other distinct roots.

OpenStudy (anonymous):

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

Parth (parthkohli):

Complex numbers kick in: \( \color{Black}{\Rightarrow x^2 = -25 \Longrightarrow x = 5i }\)

Parth (parthkohli):

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

OpenStudy (anonymous):

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..

Parth (parthkohli):

You may find the square root of -9: \( \color{Black}{\Rightarrow \sqrt{-9} = \sqrt{-1} \times \sqrt{9} = i \times 3 = 3i}\)

Parth (parthkohli):

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

Parth (parthkohli):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Are you getting what parth has done @Steeben ??

OpenStudy (anonymous):

@ 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..

Parth (parthkohli):

No!

Parth (parthkohli):

The equation has only one root with multiplicity 2.

OpenStudy (anonymous):

I can prove my point..

OpenStudy (anonymous):

okay guys... no need to fight, ... BE HAPPY LOL k so I started with this:

Parth (parthkohli):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

x^2+4x+13=0

OpenStudy (anonymous):

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

OpenStudy (anonymous):

Yes you are on a right track @Steeben

OpenStudy (anonymous):

(x+2)^2=-9

OpenStudy (anonymous):

No this one is not right..

OpenStudy (anonymous):

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

Parth (parthkohli):

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

OpenStudy (anonymous):

See, \(x^2 + 4x + 13 = 0\) You got this one??

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

@ParthKohli how did you get a random variable i?

OpenStudy (anonymous):

@ParthKohli please let me explain my point..

Parth (parthkohli):

\( \color{Black}{\Rightarrow x^2 + 4x = -13 }\) \( \color{Black}{\Rightarrow x^2 + 4x + 4 = -13 + 4}\) \( \color{Black}{\Rightarrow (x + 2)^2 = -9}\) That is the complex number \(i \).

Parth (parthkohli):

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

OpenStudy (anonymous):

@Steeben I am telling you the correct way to do this.. Can you listen to me or not??

OpenStudy (anonymous):

okay just type it out

OpenStudy (anonymous):

Using quadratic formula: \(D = b^2 - 4ac\) D = 16 - 52 = -36 \[x = \frac{-4 \pm \sqrt{-36i}}{2}\] See, \(\sqrt{-36} = \sqrt{(-1) \times 36} = 6i\)

OpenStudy (anonymous):

Put in the equation and solve: \[x = -2 \pm 3i\] So this has two solutions: \(x = -2 + 3i\) and \(x = -2 - 3i\)

OpenStudy (anonymous):

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

Parth (parthkohli):

Okay. You win, I lose.

OpenStudy (anonymous):

That is better.. Ha ha ha..

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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.

OpenStudy (anonymous):

You want to know about i??

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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??

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so i is kinda "replacing" the negative?

OpenStudy (anonymous):

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

OpenStudy (anonymous):

12i

OpenStudy (anonymous):

Yes nice.. You are learning very fast..

OpenStudy (anonymous):

I show you why \(\sqrt{-144} = 12i\) See first you have to remember that" \[\sqrt{ab} = \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\]

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

|dw:1342805660036:dw|

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!