1. ## 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?

2. ## Re: Solving a Disguised Quadratic

Originally Posted by Luminary
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?

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.

3. ## 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.