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Thread: Algebra

  1. #1
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    Algebra

    Q1= show that if x is real, 2(x^2) + 6x + 9 is always positive.

    Q2= Solve the simultaneous equations for x,y > 0

    2log y = log 2 + log x

    2^y = 4^x

    cheers
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  2. #2
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    for Q1 check the discriminant of the quadratic $\displaystyle b^2 -4ac$

    for Q2 using the first equation you should be able to get $\displaystyle \log y^2 = \log 2x \Rightarrow y^2 = 2x$ using the laws of logs.

    log both sides on the second equation and it should be easy form there.
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  3. #3
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    Quote Originally Posted by sparky69er View Post
    Q1= show that if x is real, 2(x^2) + 6x + 9 is always positive.

    Q2= Solve the simultaneous equations for x,y > 0

    2log y = log 2 + log x

    2^y = 4^x

    cheers
    An alternative "looking" way to show Q1 would be completing squares and examining the expression.
    $\displaystyle
    2(x^2 + 3x + \frac92) = 2([x^2 + 2.\frac32.x + (\frac32)^2] + \frac92 - (\frac32)^2) = 2(x + \frac32)^2 + \frac94$

    Now since $\displaystyle (x + \frac32)^2$ is always greater than or equal to zero, for all real x, and since $\displaystyle \frac94$ is positive , its sum is strictly positive.
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  4. #4
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    Quote Originally Posted by bobak View Post
    for Q1 check the discriminant of the quadratic $\displaystyle b^2 -4ac$

    for Q2 using the first equation you should be able to get $\displaystyle \log y^2 = \log 2x \Rightarrow y^2 = 2x$ using the laws of logs.

    log both sides on the second equation and it should be easy form there.
    Hi, for Q1 i have done as you said and got -36 but what do you mean by the discriminant????
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  5. #5
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    Hello, sparky69er!

    1) Show that if $\displaystyle x$ is real, $\displaystyle 2x^2 + 6x + 9$ is always positive.

    We have: .$\displaystyle x^2 + x^2 + 6x + 9 \:=\:x^2 + (x+3)^2$

    The sum of two squares is always nonnegative.

    When $\displaystyle x = 0$, the polynomial has a minimum value of 9.



    2) Solve the simultaneous equations for x, y > 0

    . . $\displaystyle \begin{array}{cc}2\log y \:=\: \log 2 + \log x & {\color{blue}[1]} \\ 2^y \: =\: 4^x & {\color{blue}[2]} \end{array}$

    Equation [1] is: .$\displaystyle 2\log y \:=\:\log(2x)\;\;{\color{blue}[3]}$


    From Equation [2]: .$\displaystyle 2^y \:=\:(2^2)^x \:=\:2^{2x}\quad\Rightarrow\quad y \:=\:2x\;\;{\color{blue}[4]}$

    Substitute into [3]: .$\displaystyle 2\log(2x) \:=\:\log(2x)\quad\Rightarrow\quad\log(2x) \:=\:0$

    . . Hence: .$\displaystyle 2x \:=\:1\quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{2}}$

    Substitute into [4]: .$\displaystyle y \:=\:2\left(\frac{1}{2}\right) \quad\Rightarrow\quad\boxed{ y \:=\:1}$

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  6. #6
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    Quote Originally Posted by sparky69er View Post
    Hi, for Q1 i have done as you said and got -36 but what do you mean by the discriminant????
    Given a quadratic polynomial equation of the form
    $\displaystyle ax^2 + bx + c = 0$,
    we have solutions
    $\displaystyle x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
    (This is the "quadratic formula.")

    The "discriminant" is defined as $\displaystyle D = b^2 - 4ac$. We have three different classes of solution based on the value of the discriminant:
    $\displaystyle D > 0 \implies$ two real, unequal roots.

    $\displaystyle D = 0 \implies$ one real root. (Or two equal real roots, however you wish to look at it.)

    $\displaystyle D < 0 \implies$ two complex conjugate roots.

    In the case of D < 0 the curve $\displaystyle y = ax^2 + bx + c$ never crosses the x axis. If a > 0 then this would imply that the quadratic is always positive.

    -Dan
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  7. #7
    Lord of certain Rings
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    Quote Originally Posted by Soroban View Post
    Hello, sparky69er!

    We have: .$\displaystyle x^2 + x^2 + 6x + 9 \:=\:x^2 + (x+3)^2$

    The sum of two squares is always nonnegative.
    Smart, really smart solution
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