Complex Numbers : Roots of a quadratic equation - conjugate pairs : ExamSolutions

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Tutorial on complex numbers. I show you how to find the roots of a quadratic equation (conjugate pairs).

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@DevilVu Have you looked at my first video on complex numbers. It explains it there.

ExamSolutions_Maths

Because if this is equal to zero then x=a or x=b. This was the line before where I have the roots.

ExamSolutions_Maths

You didn’t explain the faster way of solving these questions which is adding the roots and multiplying them out

kiyodante

Your Voice, Your Accent... <3 Stay Blessed Sir. Thanks.

angelinearon

Sub x=3-i into the equation and then equate real and imag. parts. to find a, and b. For the other root you could then sub x=ki into the equation and compare real and imag. parts

ExamSolutions_Maths

isnt there a simpler way of working it out in the text book?

takmaps

where the hell did you get i from mate? 1:57

abbaibrahim

Sir, why, (x-a)(x-b) ? is it because, this is how quadratic equation is formed ? also why both signs are negative ?

emad

Why is (x-α)(x-β) = 0 implied? I've seen you solve for the roots, before. Couldn't you also just use (x+α)(x-β) = 0 or (x+α)(x+β) = 0? I know two minus signs would make it pretty easy to solve. Do these other forms somehow not always give rise to a quadratic equation?

topaussiemezza

thank you sir i am from india (bharat )

HeyGRArnab

ExamSolutions
Hi sir, may I ask you since the equation takes the form of (x-a)(x-b), surely if x=a then x=b and hence a=b too right? I am still unsure why the equation is (x-a)(x-b).

And finally sir, for both questions, I worked out a+b and ab and then subbed the values in the equation x^2 - (a+b)x +ab, by expanding (x-a)(x-b) - this gets the same answer as you did, so is it acceptable to use either method.

Regards sir!

jayagnihotri

Sir when you multiply -3i and 3i, why do you get 9

emmanuellul

You should have just used sum/product approach as one would with surd conjugates.

trevorsimpson

Since i^2 = -1 and -1 x 16 = -16, that means -16 can be simplified to 16i^2. If you take the square root of this, 16 square rooted is 4 (or -4) and i^2 square rooted is i. So, the roots of -16 are 4i and -4i.

Vinnyz

What about if there is a 3rd root we have to find, or is this only in cubics? This is useful to know, and when i factorise the quadratic at the end it leads me back to the same roots of the original.

iDontcare_

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