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Math Help - Solving equations

  1. #1
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    Solving equations

    Solve each equation by the most suitable method. please do not use decimals in the answer.



    b)

    z(5z-1)=3z(z+3)

    c)

    x^2+6x-8=0

    d)

    x^2 +8x/3=1
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  2. #2
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    Quote Originally Posted by pashah
    Solve each equation by the most suitable method. please do not use decimals in the answer.
    b) z(5z-1)=3z(z+3)
    c) x^2+6x-8=0
    d) x^2 +8x/3=1
    Hello, pashah,

    to b): rearrange, so that the RHS of the equation is zero. Put the factor z befor the bracket. Then you have two factors. This product can be only zero if one of the factors is zero:
    z(5z-1)-3z(z+3)=0\Longleftrightarrow z(5z-1-3z-9)=0 \Longleftrightarrow z(2z-10)=0 \Longleftrightarrow z=0\ \vee\ z=5

    to c) (Are you sure that there isn't a typo? If you have +8 instead of -8 then you could use factorization) Use the formula:
    x_{1,2}=-\frac{6}{2\cdot 1}\pm\sqrt{\frac{6^2-4\cdot 1\cdot (-8)}{4\cdot 1^2}} \  \quad \Longleftrightarrow\ \ x=-3+\sqrt{17}\vee x=-3-\sqrt{17}

    to d) Use the formula:
    x_{1,2}=-\frac{\frac{8}{3}}{2}\pm\sqrt{\frac{\frac{64}{9}-4\cdot 1\cdot (-1)}{4\cdot 1^2}} \Longleftrightarrow x=-\frac{4}{3}+\frac{5}{3}=\frac{1}3}\vee x=-\frac{4}{3}-\frac{5}{3}=-3

    Greetings

    EB
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  3. #3
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    Hello, pashah!

    I assume you know how to solve a quadratic equation.
    . . Get all terms on the left (0 on the right).
    . . Factor the quadratic (if possible).
    . . Set each factor equal to zero and solve the resulting equations.


    b)\;\;z(5z-1)\:=\:3z(z+3)

    Distribute: . 5z^2 - z\;=\;3x^2 + 9z

    We have: . 2z^2 - 10z\;=\;0

    Factor: . 2z(z - 5)\;=\;0

    And we have: . \begin{array}{cc}2z\:=\:0\quad\Rightarrow\quad \boxed{z = 0} \\ z - 5\:=\:0\quad\Rightarrow\quad \boxed{z = 5}\end{array}



    c)\;\;x^2 + 6x - 8\:=\:0

    I agree with EB . . . I'm certain there's a typo.

    Suppose it was: . x^2 + 6x + 8 \:=\:0

    Factor: . (x + 2)(x + 4)\:=\:0

    Then we have: . \begin{array}{cc}x+2\,=\,0\quad\Rightarrow\quad \boxed{x = -2}\\x + 4 \,=\,0\quad\Rightarrow\quad \boxed{x = -4}\end{array}



    d)\;\;x^2 + \frac{8x}{3} \:=\:1

    Multiply by 3: . 3x^2 + 8x\:=\:3

    We have: . 3x^2 + 8x - 3 \:= \:0

    Factor: . (3x - 1)(x + 3)\:=\:0

    Then we have: . \begin{array}{cc}3x - 1\:=\:0\quad\Rightarrow\quad \boxed{x = \frac{1}{3}} \\ x + 3\;=\:0\quad\Rightarrow\quad \boxed{x = -3}\end{array}

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