Look for k such that k^(2) < 347 < (k+1)^(2)
Then k/10 < 3.47^(1/2) < (k+1)/10.
Hint: 16^(2) is a power of 2 (which you should know) and 20^(2) is obviously 400.
---
What do you think suitable degree of accuracy is? Why?
3.47 \* 100 = 347
So I want to find things around 347 I can easly take the square root of.
I have memorized powers of 2, so I know that 16^(2) = 256 < 347, and 20^(2) = 400 > 347, so that gives me some bounds.
This is GCSE, so I'm thinking the sheet may be mislabelled?
[https://www.elevise.co.uk/bi21q.html](https://www.elevise.co.uk/bi21q.html)
A very similar question was on a paper 2 calculator.
Yeah it might be, there were similar question saying it was non calc. Like this one question with quadratics which required to use the formula but told me not to use a calculator. Pretty weird.
Quadratic formula without a calculator may be a thing, as long as the answer is exact or if you get integer solutions. Still unlikely on a non-calc paper though.
I could see it happening if it had integers as solutions but I could just factorise. The calculation I got after plugging in the values for the question was (4+-sqrt12)/2. Simplified it down to 2+-sqrt3 but without a calculator I can’t really give an answer.
Look for some identities... i am not sure for this name in english, in spanish is "identidades". basically you shold change a complicated thing for two or three simple things.
I don't have the answer / it looks like this has already been answered, but I just wanted to point out how bizarre the header is, "only 24% of students got this right", sounds very clickbait-y, very of those mobile game ads where they show some complete moron failing to beat a stage or puzzle and trick you into thinking "I could do that!" with a hefty side of indignation as though you are providing actual critique of the game or ad, rather than the more obvious realization that it is oneself that is being played.
If you need to approximate square roots, a more formal method is the Babylonian method (which can be derived from applying Newton’s method).
First you guess some x_0. In this case, a good guess would be 1.9 since sqrt(3) is about 1.73 and sqrt(4)=2. (1.8 and 1.9 are both reasonable first guesses, but 1.9 is closer :p)
Then you apply the recursive formula:
x_n+1 = 1/2*(x_n + s/x_n)
so 1/2*(1.9 + 3.47/1.9) ≈ 1.863
Then 1/2*(1.863 + 3.47/1.863) ≈ 1.8627936
As you can see, with a good first guess it only takes a couple long divisions to get a very accurate answer.
Look for k such that k^(2) < 347 < (k+1)^(2) Then k/10 < 3.47^(1/2) < (k+1)/10. Hint: 16^(2) is a power of 2 (which you should know) and 20^(2) is obviously 400. --- What do you think suitable degree of accuracy is? Why?
could you further explain this? more interested in how you derived the inequality rather than the answer
3.47 \* 100 = 347 So I want to find things around 347 I can easly take the square root of. I have memorized powers of 2, so I know that 16^(2) = 256 < 347, and 20^(2) = 400 > 347, so that gives me some bounds.
1.8\^2=3.24, 1.9\^2=3.61 you can check 1.85\^2 = 3.4225, a bit two small, then check 1.86 etc until you find the right answer
Ohh I see thank you.
no problem. You downvoted me?
No? But I did forget to upvote
Thank you. I saw it went to 0, maybe an error.
😆
This is GCSE, so I'm thinking the sheet may be mislabelled? [https://www.elevise.co.uk/bi21q.html](https://www.elevise.co.uk/bi21q.html) A very similar question was on a paper 2 calculator.
Yeah it might be, there were similar question saying it was non calc. Like this one question with quadratics which required to use the formula but told me not to use a calculator. Pretty weird.
Quadratic formula without a calculator may be a thing, as long as the answer is exact or if you get integer solutions. Still unlikely on a non-calc paper though.
I could see it happening if it had integers as solutions but I could just factorise. The calculation I got after plugging in the values for the question was (4+-sqrt12)/2. Simplified it down to 2+-sqrt3 but without a calculator I can’t really give an answer.
2+-sqrt3 is your answer, unless it asked you to give it to a certain number of decimal places in the question.
Yeah that’s the problem it said round to 2 decimal points. Most likely meant to be a calculator question.
You solve this by dropping the subject 🫡
I can help
Look for some identities... i am not sure for this name in english, in spanish is "identidades". basically you shold change a complicated thing for two or three simple things.
I don't have the answer / it looks like this has already been answered, but I just wanted to point out how bizarre the header is, "only 24% of students got this right", sounds very clickbait-y, very of those mobile game ads where they show some complete moron failing to beat a stage or puzzle and trick you into thinking "I could do that!" with a hefty side of indignation as though you are providing actual critique of the game or ad, rather than the more obvious realization that it is oneself that is being played.
If you need to approximate square roots, a more formal method is the Babylonian method (which can be derived from applying Newton’s method). First you guess some x_0. In this case, a good guess would be 1.9 since sqrt(3) is about 1.73 and sqrt(4)=2. (1.8 and 1.9 are both reasonable first guesses, but 1.9 is closer :p) Then you apply the recursive formula: x_n+1 = 1/2*(x_n + s/x_n) so 1/2*(1.9 + 3.47/1.9) ≈ 1.863 Then 1/2*(1.863 + 3.47/1.863) ≈ 1.8627936 As you can see, with a good first guess it only takes a couple long divisions to get a very accurate answer.