How to solve quadratic equations by using the quadratic formula

In this algebra tutorial, we will solve two quadratic equations by using the quadratic formula. We have already solved these two equations by completing the square and you can check out the video here: • Solving quadratic equa...
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#math #algebra #mathbasics

Пікірлер: 22

  • @bprpmathbasics
    @bprpmathbasics2 ай бұрын

    Solve them by completing the square: kzread.info/dash/bejne/qYppr86caNPXYLA.htmlsi=qBFAqbDKLCgHE_Ka

  • @ASHHH_Art

    @ASHHH_Art

    Ай бұрын

    ok :)

  • @ASHHH_Art
    @ASHHH_ArtАй бұрын

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

  • @hannah-yn2fh
    @hannah-yn2fh4 күн бұрын

    THANK YOU

  • @bodesshorts8640
    @bodesshorts8640Ай бұрын

    I wish this video was before my final exam

  • @buttonmasherbaberkins7490
    @buttonmasherbaberkins7490Ай бұрын

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

  • @sirhumv33
    @sirhumv33Ай бұрын

    9:21 that stutter

  • @diewand5442
    @diewand5442Ай бұрын

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

  • @carultch

    @carultch

    Ай бұрын

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

  • @diewand5442

    @diewand5442

    Ай бұрын

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

  • @teelo12000
    @teelo12000Ай бұрын

    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

    @carultch

    Ай бұрын

    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

    @andrewm6424

    Ай бұрын

    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.”

  • @JeePrepration
    @JeePreprationАй бұрын

    It is not quadratic fourmula it is shree dharacharya fourmula

  • @davidwebster9788
    @davidwebster9788Ай бұрын

    Why not do it this way all the time?

  • @LaMirah

    @LaMirah

    Ай бұрын

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

  • @davidwebster9788

    @davidwebster9788

    Ай бұрын

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

  • @tobybartels8426

    @tobybartels8426

    Ай бұрын

    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.

  • @zachansen8293
    @zachansen8293Ай бұрын

    too basic

  • @Noobman69420

    @Noobman69420

    Ай бұрын

    Read the channel name

  • @zachansen8293
    @zachansen8293Ай бұрын

    too basic

  • @tigerlover7359

    @tigerlover7359

    Ай бұрын

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