How to solve quadratic equations by using the quadratic formula

[@bprpmathbasics]

Solve them by completing the square: kzbin.info/www/bejne/rYeYfp-ZbdWkZq8si=qBFAqbDKLCgHE_Ka

[@ASHHH_Art]

ok :)

[@ASHHH_Art]

I love watching u😁, also I have my JEE exam.

[@bodesshorts8640]

I wish this video was before my final exam

[@hannah-yn2fh]

THANK YOU

[@HUMV33]

9:21 that stutter

[@rivenoak]

3 and 19 are primes both, the sqare roots will be a irrationals and multiplication of irrational numbers wont help either

[@Musterkartoffel]

I'll stay with my pq-formula, but still thanks

[@carultch]

How about the "m +/- sqrt(m^2 - p)" formula, for mean-product?

[@Musterkartoffel]

@@carultch true, I forgot there's a 3rd way, I always forget it exists 😅

[@buttonmasherbaberkins7490]

I'm having trouble understanding quadratics, in application. I get the linear part, it's x^2 that I don't quite get. 😢

[@teelo12000]

Seems poor phrasing to call them "not factorable" only to use the formula to get their factors, no? First one factors to (x-5+sqrt(21))(x-5-sqrt(21))=0

[@carultch]

I think "not factorable in rational numbers" is what he really meant. For applications of this concept, you'd be correct. If it is possible to factor to real solutions, you'd still want to do so, even if the factors are irrational. For instance, given the integral 2*sqrt(21)/(x^2 - 10 x + 4) dx, you'd still prefer to first factor it to 2*sqrt(21)/((x - 5 + sqrt(21))*(x - 5 - sqrt(21))), rather than following the procedure for an "irreducible quadratic". The procedure for an irreducible quadratic, will expect a denominator of ((x - a)^2 + b^2), with b^2 being positive, and produces a different function family entirely, than what you'll want for this one. You'd then construct partial fractions; 1/(x - sqrt(21) - 5) - 1/(x + sqrt(21) - 5) And integrate as: ln(|x - sqrt(21) - 5|) - ln(|x + sqrt(21) - 5|) + C

[@andrewm6424]

Good point. But by the time you learn the quadratic formula, you’re expected to know that “not factorable” means “not able to get your factors by ‘FOIL’ method.”

[@davidwebster9788]

Why not do it this way all the time?

[@LaMirah]

That's what many of us do in practice, especially in applications where you're not guaranteed real roots.

[@davidwebster9788]

@@LaMirah Seems better in the end as there are no algorithms to remember.

[@tobybartels8426]

If it's obvious how to factor it, then that's usually faster, and has less room for arithmetic errors. But if it's not obvious how to factor it, then it's probably not worth the trouble to keep trying or to verify that you can't, so use the formula. As for completing the square, that's usually less convenient; however, it can be used for other things (such as finding equations of parabolas and other conic sections), so it's still useful to know how to do it.

[@JeePrepration]

It is not quadratic fourmula it is shree dharacharya fourmula

[@zachansen8293]

too basic

[@Noobman69420]

Read the channel name

[@zachansen8293]

too basic

[@tigerlover7359]

It’s as if this particular channel is about math basics. Unless you can’t read which is clearly the case.