Results 1 to 3 of 3

Math Help - help pls

  1. #1
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261

    help pls

    Solve the following equations .

    (1) \sqrt{x-2}+\sqrt{x+3}=5


    (2) \sqrt{2x+11}+\sqrt{x+2}=\sqrt{5x+17}
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Dec 2008
    From
    India
    Posts
    7
    1.sqrt(x-2)+sqrt(x+3)=5
    squaring both thr sides we have(sqrt(x-2)+sqrt(x+3))^2=25
    using (a+b)^2 formula i.e a^2+ b^2+2ab here we get
    x-2 +x+3 +2(sqrt(x-2)sqrt(x+3)0=25
    2x+1-25 = -2sqrt(x-2)sqrt(x+3)
    2(x-12)=-2sqrt(x-2)sqrt(x+3)
    x-6 = -sqrt(x-2)sqrt(x+3)
    taking square both the sides we have
    (x-6)^2 = (-sqrt(x-2)sqrt(x+3))^2
    x^2-12x+36 = (x-2)(x+3)
    x^2-12x+36= x^2+x-6
    taking the variables to the l.h.s and constants to r.h.s we have
    -12x-x=-36-6
    -13x = -42
    -13x/-13= -42/-13
    so,x =4
    the next sum you can try similarly
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Sep 2008
    From
    West Malaysia
    Posts
    1,261
    Quote Originally Posted by sunitaparija View Post
    1.sqrt(x-2)+sqrt(x+3)=5
    squaring both thr sides we have(sqrt(x-2)+sqrt(x+3))^2=25
    using (a+b)^2 formula i.e a^2+ b^2+2ab here we get
    x-2 +x+3 +2(sqrt(x-2)sqrt(x+3)0=25
    2x+1-25 = -2sqrt(x-2)sqrt(x+3)
    2(x-12)=-2sqrt(x-2)sqrt(x+3)
    x-6 = -sqrt(x-2)sqrt(x+3)
    taking square both the sides we have
    (x-6)^2 = (-sqrt(x-2)sqrt(x+3))^2
    x^2-12x+36 = (x-2)(x+3)
    x^2-12x+36= x^2+x-6
    taking the variables to the l.h.s and constants to r.h.s we have
    -12x-x=-36-6
    -13x = -42
    -13x/-13= -42/-13
    so,x =4
    the next sum you can try similarly

    I guess you made a mistake somewhere
    \sqrt{x-2}+\sqrt{x+3}=5
    2x+1+2\sqrt{x-2}\sqrt{x+3}=25
    (-x+12)^2=(2\sqrt{x-2}\sqrt{x+3})^2
    144-24x+x^2=(x-2)(x+3)
    and i finally got x=6
    however , thanls for helping me out .
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum