solving 1/2 + 1/sqrt(x) = 1/3 but you won't get a solution

Rational equation with square root.

Пікірлер: 23

  • @ShubhayooBanerjee
    @ShubhayooBanerjee5 ай бұрын

    Eminem been real quiet since this banger dropped🗣🗣🔥🔥

  • @manthankashyap933
    @manthankashyap9335 ай бұрын

    The answer is 1/sqrt(x)=-1/6, But, there is no value of x, which satisfies this equation

  • @aneeshbro
    @aneeshbro5 ай бұрын

    Nah bro be rappin' math 🗿🗿🗿🗿

  • @syndrac6254
    @syndrac62545 ай бұрын

    When your math teacher is Eminem. Lyrics coming at you at super sonic speed.

  • @rythmicconvival
    @rythmicconvival5 ай бұрын

    Nice

  • @spthepero2282
    @spthepero22825 ай бұрын

    BUT WHY TF? IF YOU PUT -6 IN PLACE OF SQRT(X), THEN YOU GET BOTH SIDES CORRECT.

  • @oZqdiac

    @oZqdiac

    5 ай бұрын

    But sqrt(x) can’t equal -6

  • @spthepero2282

    @spthepero2282

    5 ай бұрын

    @@oZqdiac so is it an extraneaous root? Oh I get it now, Ik about extraneaous root stuff

  • @bhaveshsinghbisht

    @bhaveshsinghbisht

    4 ай бұрын

    ​@@spthepero2282 actually 36 is not a solution . When you will put it in equation. , it will not satisfy .Let me tell you why 1/2+1/√36=1/2+1/6=4/6=2/3 | | \ / You took √36=-6 which is wrong . √36=6 only . That was the mistake and you got your answer correct . Hope this helps❤.

  • @spthepero2282

    @spthepero2282

    4 ай бұрын

    @@bhaveshsinghbisht Yeah thanks for the clarifaction mate, I didn't realise that was an extraneous root.

  • @Happylittleradish
    @Happylittleradish4 ай бұрын

    Why dident you rationalize the denominator?

  • @f.r.y5857
    @f.r.y58575 ай бұрын

    √x² = |x|

  • @teoteo8351
    @teoteo83515 ай бұрын

    Apologies for a completely unrelated question. I am currently studying University, completed the chapters for basic differentiation and integration, what should I learn next or what can be my to-learn list? The professor won't upload any materials for me to prepare...

  • @user-gd1ej9eh1k

    @user-gd1ej9eh1k

    5 ай бұрын

    Linear algebra and real analysis

  • @madsorcery

    @madsorcery

    5 ай бұрын

    Complex Numbers.

  • @johnathanpatrick6118
    @johnathanpatrick61185 ай бұрын

    Just looking at the equation itself you can tell 1/sqrt(x) had to somehow be negative since 1/2 > 1/3. Never gonna work.

  • @monttdher
    @monttdher5 ай бұрын

    Talk about conic sections

  • @robzan8
    @robzan85 ай бұрын

    Does sqrt(x) = -6 have a complex solution?

  • @user-cc6ci7jx5q

    @user-cc6ci7jx5q

    5 ай бұрын

    No even the complex world can solve this as i (as the sqrt of -1) can’t be plugged anywhere

  • @allozovsky

    @allozovsky

    3 ай бұрын

    ​@@user-cc6ci7jx5q In fact, that's not that strict and mostly depends upon how we treat the square root notation. Steve himself in one of his videos (sqrt(i)) claimed that "√𝒊 has *two* answers", namely √𝒊 = ±(1 + 𝒊)/√2, so in the same manner one might claim that √(-1) = ±𝒊 and √36 = √(36 + 0·𝒊) = ±6, but that leads us to multivaluedness, which we mostly try to avoid in our notation (as much as we can).

  • @comdo777
    @comdo7775 ай бұрын

    asnwer=1 isit

  • @69Gigantosaur
    @69Gigantosaur5 ай бұрын

    No comments? 😮😮

  • @Grizzly01-vr4pn
    @Grizzly01-vr4pn5 ай бұрын

    End card obscuring pertinent parts of the video. Poor upload.