solve quadratic equation by square root property
Uploader Comments (robichaudd)
Top Comments
-
thank you so much! i understood more from watching this video than struggling with the text book for the past few hours!
-
hey girl you much be some sort of good teacher. u've helped me alot thanks alot!!!!!!!!
All Comments (42)
-
STRESSED NO MOAR, thank you :D
-
Is it me or she sounds like sara quin..?
-
Excellent lesson and very concise explanation, thanks^9000!
-
you just saved me from having a panic attack thank you
-
@TheOneAndOnlyKinable move the 7 on to the other side then do her process sorry im 10 months late :p
-
U R BEST TEACHER
-
@robichaudd ahhh yes. i got it, thanx. i had some equations using the square root property and while solving for sqrt, i would get an error on my calculator. this drove me nuts, i found out that i have to factor out an (i) for the negative sqrts. we covered this, i just got a little overwhelmed.
-
@ 3:38 there are times when your answer has an i on the end...ex: plus or minus 7i. why is that...?
-
The square root of 36 is 6
-
Ok, I have a question: I've got this problem --> (x+7)² = -36 and the next step I've been taught to take is to then have x+7 = +/- √36 but I don't know what to do with the √36. Would the equation then be x+7 =+/- 2√3?
WOW HOW DO YOU SOLVE A THREE STEP OMG EVERYONE IS DOING EASY EQUATIONS ON HERE WOW, can someone plz do x^2 + 6x - 7 = 0 by square root plzzz.
TheOneAndOnlyKinable 1 year ago
@TheOneAndOnlyKinable Search for a video on "solve quadratic equation by completing the square." At some point in the process the square root property is used. I think that might be what you're looking for.
robichaudd 1 year ago
can you do the question x^2 -2x+5=0, im kind of stuck there. I have to solve for complex roots
mrhamada56 1 year ago
@mrhamada56 Sorry but I only have time to answer follow up questions to the problems in the video. Keep searching youtube - there must be some videos out there for solving quadratic equations with complex solutions.
robichaudd 1 year ago
...I assume that "principal" square root
does not apply since these roots are
solutions of quadratic equations, not
constants (e.g., square root of 9 is 'only'
three and not also -3).
jwm239 3 years ago
That's correct. For something like x^2=9 we can also see how we need both +3 and -3 as solutions by solving it by factoring instead:
x^2=9 ... x^2-9=0 ... (x+3)(x-3)=0 ... x=-3,3
robichaudd 3 years ago