| Marsh
Narewec |
My cousin brother is having problem working out the cubic root of 125 i.e.
. Of course I know the exact answer is 5 but how can he work out using surds. Please explain. Thank you |
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| David
Chen |
How about factorise 125, then we get 5 * 5 * 5, which equals to 53 . |
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| Matthew
Smith |
I don't think that surds are a good method of finding cube roots. Perhaps you are looking for a way of expressing the cube root of a number that isn't a perfect cube (such as the cube root of 2 ) exactly in terms of surds, but this isn't possible in general. Where the number is small and it's a perfect cube, like 125, we can easily find the cube root by inspection, as David said. Where this isn't possible, we can only find an approximation. There are various ways we could go about this... 1) The easiest way is to get tables of cube roots, log tables, or a calculator. Assuming that these aren't allowed... 2) The simplest way is to guess an answer, cube it, and see if the result is greater or less than the number we're interested in. If it's greater, our guess was too big; if less, too small. We can then adjust our guess to take this into account. This way, we soon arrive at a range in which the cube root must lie. At each stage, we can narrow this range by finding out whether the number in the middle is too small or too big. This method is called bisection. 3) Another way is to start with a guess, which we call x0 , and work out by about how much it needs to be adjusted. If we call this amount d, and the number whose cube root we want is C, we have (x0 +d)3 =C. Expanding the bracket, x0 3 +3x0 2 d+3x0 d2 +d3 =C. Now if x0 is a reasonably good guess, d is quite a bit smaller than x0 , so to a good approximation we can ignore the terms with d2 and d3 . Solving the rest of the equation for d, we get d=(C-x0 3 )/(3x0 2 ), and adding this on to x0 we get a new, better guess. We can keep on doing this to get closer and closer to the cube root, and it converges quite quickly. 4) If you really want to use surds, you could use a similar method to the above, but only ignore the d3 term, so that you keep the d3 term and need to solve a quadratic for d. This would converge even faster than method (3), but you'd have to work out the values of the square roots as you went along, which would slow it down. If you left the square roots in surd form, you'd only have a long and complicated expression for an approximation to the cube root, not the cube root itself. Matthew |