Solve by completing the square. z^2=8z-16
k so what I got is, z^2-8z=-16 then z^2-8z+16=0
but then, (z-4)^2=0
but I can't find the square root of 0 right?
Yes you are right..
You are on the right track. Find the square root of both sides.
You can.
\(\sqrt0 = 0\)
oh. z-4=0?!
Yes..
Exactly :)
so z = 4 and -4 right.
Oops.
No..
Add 4 both the sides..
only one answer?
You don't need to complicate it. \( \color{Black}{\Rightarrow z - 4 + 4 = 0 + 4}\)
Yes. We have only one root in a perfect square :)
No you will get two but same answer..
oh and also, what happens when the right side of the equation is a negative? how will i do the square root then?
You may say that 4 is a root with multiplicity 2.
But there are no other distinct roots.
another question, I have (x+2)^2=-9...
Complex numbers kick in: \( \color{Black}{\Rightarrow x^2 = -25 \Longrightarrow x = 5i }\)
\( \color{Black}{\Rightarrow x + 2 = 3i }\)
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..
You may find the square root of -9: \( \color{Black}{\Rightarrow \sqrt{-9} = \sqrt{-1} \times \sqrt{9} = i \times 3 = 3i}\)
@waterineyes No need to complicate it there too. Just find the square root of both sides :)
Back to the point: \(x + 2 = 3i\) then you subtract 2 from both sides.
That is not applicable and not the correct method though @ParthKohli
Are you getting what parth has done @Steeben ??
@ 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..
No!
The equation has only one root with multiplicity 2.
I can prove my point..
okay guys... no need to fight, ... BE HAPPY LOL k so I started with this:
You may find the square root in case of a perfect square.
No.. this is complex one dear and not real..
x^2+4x+13=0
x^2+4x=-13 x^2+4x+4=-13+4
Yes you are on a right track @Steeben
(x+2)^2=-9
No this one is not right..
then then i dont' know what to do... supposedly 9 is going to be 3 but it's negative :/
I already showed how \(\sqrt{-9} = 3i\)
See, \(x^2 + 4x + 13 = 0\) You got this one??
yes
@ParthKohli how did you get a random variable i?
@ParthKohli please let me explain my point..
\( \color{Black}{\Rightarrow x^2 + 4x = -13 }\) \( \color{Black}{\Rightarrow x^2 + 4x + 4 = -13 + 4}\) \( \color{Black}{\Rightarrow (x + 2)^2 = -9}\) That is the complex number \(i \).
\(i = \sqrt{-1}\)
@Steeben I am telling you the correct way to do this.. Can you listen to me or not??
okay just type it out
Using quadratic formula: \(D = b^2 - 4ac\) D = 16 - 52 = -36 \[x = \frac{-4 \pm \sqrt{-36i}}{2}\] See, \(\sqrt{-36} = \sqrt{(-1) \times 36} = 6i\)
Put in the equation and solve: \[x = -2 \pm 3i\] So this has two solutions: \(x = -2 + 3i\) and \(x = -2 - 3i\)
@ParthKohli God has given you eyeballs check with them and tell me are they the same root with multiplicity of 2???
Okay. You win, I lose.
That is better.. Ha ha ha..
It is not about win and loss it is about knowledge..
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.
You want to know about i??
my teacher says that if there's a negative square root it's not solvable.
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??
See, \(\sqrt{36} = 6\) \(\sqrt{-36} = 6i\)
so i is kinda "replacing" the negative?
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
12i
Yes nice.. You are learning very fast..
I show you why \(\sqrt{-144} = 12i\) See first you have to remember that" \[\sqrt{ab} = \sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\]
so (x+2)^2=-9 and then x+2 = 3i
You cannot solve it directly, the correct way I have done above using Quadratic Formula..
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