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Math Help - Solving a Disguised Quadratic

  1. #1
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    Solving a Disguised Quadratic

    Hi all.

    I'm having trouble trying to solve this problem without a calculator.

    The first part of the question is to solve x^2-(6\sqrt3))x+24=0, giving your answer in terms of surds. I solved this and got 4\sqrt3 and 2\sqrt3.
    For the next part of the question, I need to solve x^4-(6\sqrt3)x^2+24=0, giving the answer correct to 2 d.p.

    By letting y=x^2, I can get that x^2=4\sqrt3 and x^2=2\sqrt3, and from there that x=\pm2\sqrt[4]3 and x=\pm\sqrt2\sqrt[4]3, but I really don't see how I could work out the values without a calculator. Any ideas?

    Thanks in advance.
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  2. #2
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    Re: Solving a Disguised Quadratic

    Quote Originally Posted by Luminary View Post
    Hi all.

    I'm having trouble trying to solve this problem without a calculator.

    The first part of the question is to solve x^2-(6\sqrt3))x+24=0, giving your answer in terms of surds. I solved this and got 4\sqrt3 and 2\sqrt3.
    For the next part of the question, I need to solve x^4-(6\sqrt3)x^2+24=0, giving the answer correct to 2 d.p.

    By letting y=x^2, I can get that x^2=4\sqrt3 and x^2=2\sqrt3, and from there that x=\pm2\sqrt[4]3 and x=\pm\sqrt2\sqrt[4]3, but I really don't see how I could work out the values without a calculator. Any ideas?

    Thanks in advance.
    Use some method of successive approximations. The crudest form is successive bifurcation.

    $x = \pm 2\sqrt[4]{3} \implies x^4 = 48 \implies 2^4 = 16 < x^4 < 81 = 3^4.$

    Let's try $2.5^4 = 6.25^2 < 6.3^2 = 39.69.$ Way too low.

    Let's try $2.7^4 = 7.29^2 > 49.$ Too high.

    Let's try $2.6^4 = 6.76^2 < 6.8^2 = 46.24.$ Too low.

    Let's try $2.65^4 = 7.0225^2 > 49.$ Too high.

    Let's try $2.64^4 = 6.9696^2 \approx 48.57.\ 48.57 - 48 = 0.57.$ Bit too high.

    Let's try $2.63^2 = 6.9169^2 \approx 47.84.\ 48 - 47.84 = 0.16.\ and\ 0.16 < 0.57.$ Bit too low, but closer than 2.64.

    $x \approx \pm 2.63.$ Using a calculator, you can check that this process gives you an answer correct to two decimal places.

    So $2\sqrt{3} \approx 2.63 \implies \sqrt{3} \approx 1.315.$

    So what is a good approximation of $\sqrt{2}?$

    $1 = 1^2 < 2 < 2^2 = 4 \implies 1 < \sqrt{2} < 2.$

    $1.5^2 = 2.25.$ Too high.

    $1.4^2 = 1.96.$ Bit too low.

    $1.43^2 = 2.0449.$ Bit too high.

    $1.42^2 = 2.0164.$ Very close, but perhaps 1.41^2 is closer. $2.0164 - 2 = 0.0164.$

    $1.41^2 = 1.9881.\ 2 - 1.9881 = 0.0119 < 0.0164.$

    $\sqrt{2} \approx 1.41 \implies \sqrt{2} * \sqrt{3} \approx 1.41 * 1.315 \approx 1.85.$

    $x = \pm 1.85.$

    There are faster methods such as Newton-Raphson, but that method requires calculus.
    Thanks from Luminary
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  3. #3
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    Re: Solving a Disguised Quadratic

    Perhaps that is what you're expected to do, although it seems like rather a lot of calculation. The 1.85 is also out by 0.01, so then you'd have to go to three decimal places in your approximations for \sqrt2 and \sqrt[4]3. I'm becoming more and more of the opinion that this question just doesn't belong in a non-calculator exercise. Thanks for your input.
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