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Math Help - Finding a question related to 'finding the square' difficult

  1. #1
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    Finding a question related to 'finding the square' difficult

    I'm currently partaking a distance learning course in A-level mathematics, and have reached chapter 4 in Pure Core Maths 1&2. Found everything pretty straight-forward so far, yet this question I would greatly appreciate help with.

    Express 3x - 12x + 5 in the form A(x - B) - C :

    From the first part of the question I have : 3(x - 2) -7

    I'm then asked to find the minimum value of each of the following expressions:

    1. 3x - 12x + 5 = -7 (The value of x obviously being 2).

    And 2. (3x - 12x +5)

    It's part 2 that's confusing me? Really unsure of where to start.
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    Re: Finding a question related to 'finding the square' difficult

    Quote Originally Posted by DonGorgon View Post
    I'm currently partaking a distance learning course in A-level mathematics, and have reached chapter 4 in Pure Core Maths 1&2. Found everything pretty straight-forward so far, yet this question I would greatly appreciate help with.

    Express 3x - 12x + 5 in the form A(x - B) - C :

    From the first part of the question I have : 3(x - 2) -7

    I'm then asked to find the minimum value of each of the following expressions:

    1. 3x - 12x + 5 = -7 (The value of x obviously being 2).

    And 2. (3x - 12x +5)

    It's part 2 that's confusing me? Really unsure of where to start.
    Are you allowed to use Calculus?
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    Re: Finding a question related to 'finding the square' difficult

    I haven't covered any calculus yet nope.
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    Re: Finding a question related to 'finding the square' difficult

    Quote Originally Posted by DonGorgon View Post
    I haven't covered any calculus yet nope.
    Hm, well first notice that \displaystyle \begin{align*} \left(3x^2 - 12x + 5\right)^2 \end{align*}, being a square, can never be negative. So it makes sense that the minimum value would be 0. Let's try to evaluate where the function will be 0...

    \displaystyle \begin{align*} \left(3x^2 - 12x + 5\right)^2 &= 0 \\ 3x^2 - 12x + 5 &= 0 \\ x &= \frac{12 \pm \sqrt{(-12)^2 - 4(3)(5)}}{2(3)} \\ x &= \frac{12 \pm \sqrt{144 - 60}}{6} \\ x &= \frac{12 \pm \sqrt{84}}{6} \\ x &= \frac{12 \pm 2\sqrt{21}}{6} \\ x &= \frac{6 \pm \sqrt{21}}{3}\end{align*}

    So the minimum value is 0, which occurs when \displaystyle \begin{align*} x = \frac{6 - \sqrt{21}}{3} \end{align*} or \displaystyle \begin{align*} \frac{6 + \sqrt{21}}{3} \end{align*}
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    Re: Finding a question related to 'finding the square' difficult

    Quote Originally Posted by Prove It View Post
    Hm, well first notice that \displaystyle \begin{align*} \left(3x^2 - 12x + 5\right)^2 \end{align*}, being a square, can never be negative. So it makes sense that the minimum value would be 0. Let's try to evaluate where the function will be 0...

    \displaystyle \begin{align*} \left(3x^2 - 12x + 5\right)^2 &= 0 \\ 3x^2 - 12x + 5 &= 0 \\ x &= \frac{12 \pm \sqrt{(-12)^2 - 4(3)(5)}}{2(3)} \\ x &= \frac{12 \pm \sqrt{144 - 60}}{6} \\ x &= \frac{12 \pm \sqrt{84}}{6} \\ x &= \frac{12 \pm 2\sqrt{21}}{6} \\ x &= \frac{6 \pm \sqrt{21}}{3}\end{align*}

    So the minimum value is 0, which occurs when \displaystyle \begin{align*} x = \frac{6 - \sqrt{21}}{3} \end{align*} or \displaystyle \begin{align*} \frac{6 + \sqrt{21}}{3} \end{align*}
    Thanks for getting back to me. Ah ok, from my attempt I used the expression in the form A(x-B) - C :

    3(x - 2) - 7 = 0

    3(x - 2) = 7

    3(x - 2) = √7

    (x - 2) = √7 / 3

    x = 2 √7 / 3
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    Re: Finding a question related to 'finding the square' difficult

    Quote Originally Posted by DonGorgon View Post
    Thanks for getting back to me. Ah ok, from my attempt I used the expression in the form A(x-B) - C :

    3(x - 2) - 7 = 0

    3(x - 2) = 7

    3(x - 2) = √7

    (x - 2) = √7 / 3

    x = 2 √7 / 3
    By the order of operations, exponentiation is done before multiplication. So when you go in reverse and solve for x, you need to undo the multiplication first.
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