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    I am doing a non calculator paper and have been asked to solve a quadratic equation to 3dp

    This involves using the quadratic formula, I have to take the square root of 109 WITHOUT a calculator
    how do I do this? do I use the rules of surds?
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    Let's call it x, just for shorthand.
    100 < x^2 < 121, so 10 < x < 11.

    109 is pretty close to half way between, so let's try 10.5. 10.5^2 = 110.25.
    As x^2 < 110.25, x < 10.5.

    Let's try 10.4. 10.4^2 = 108.16.
    As x^2 > 108.16, x > 10.4.

    So now we have that 10.4 < x < 10.5. Keep trying new numbers (after this step, try something between 10.4 and 10.5) and squaring them, and you'll get closer to the right answer each time. I'm surprised it's asking it to 3dp though.
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    (Original post by starkrush)
    Let's call it x, just for shorthand.
    100 < x^2 < 121, so 10 < x < 11.

    109 is pretty close to half way between, so let's try 10.5. 10.5^2 = 110.25.
    As x^2 < 110.25, x < 10.5.

    Let's try 10.4. 10.4^2 = 108.15.
    As x^2 > 108.15, x > 10.4.

    So now we have that 10.4 < x < 10.5. Keep trying new numbers (after this step, try something between 10.4 and 10.5) and squaring them, and you'll get closer to the right answer each time. I'm surprised it's asking it to 3dp though.
    edexcel for ya ...

    haha
    thanks bro
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    (Original post by madfish)
    I am doing a non calculator paper and have been asked to solve a quadratic equation to 3dp

    This involves using the quadratic formula, I have to take the square root of 109 WITHOUT a calculator
    how do I do this? do I use the rules of surds?
    You will not be asked this on C1
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    (Original post by TenOfThem)
    You will not be asked this on C1
    for some reason it's in a C1 section of my text book with pastpaper questions and at the top it states "do not use a calculator" :mad:

    Haha, is it okay If I ask you something TenOfThem?

    This question:"by completing the square, find in terms of k the roots of the equation x^2 +2kx - 7 = 0

    okay so I have completed the square to get: (x+k)^2 -k^2 -7 = 0

    what am I meant to do next?
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    (Original post by madfish)
    for some reason it's in a C1 section of my text book with pastpaper questions and at the top it states "do not use a calculator" :mad:

    Haha, is it okay If I ask you something TenOfThem?

    This question:"by completing the square, find in terms of k the roots of the equation x^2 +2kx - 7 = 0

    okay so I have completed the square to get: (x+k)^2 -k^2 -7 = 0

    what am I meant to do next?
    Solve that for x

    eg

    (x-3)^2 - 12 = 0

    gives

    (x-3) = \sqrt{12}

    gives

    x = 3 \pm \sqrt{12}
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    (Original post by madfish)
    for some reason it's in a C1 section of my text book with pastpaper questions and at the top it states "do not use a calculator"
    old questions from when P1 was calculator
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    (Original post by TenOfThem)
    old questions from when P1 was calculator
    thanks tenofthem!

    dont know what I would do without you, If I happen to do well in my ASlevel maths I owe you big time!
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    (Original post by madfish)
    I am doing a non calculator paper and have been asked to solve a quadratic equation to 3dp

    This involves using the quadratic formula, I have to take the square root of 109 WITHOUT a calculator
    how do I do this? do I use the rules of surds?
    You use long-square-rooting, the same way as you do long division in primary school.

    Write down your number with a leading zero and some extra zeros after the decimal point because we need to deal with the digits in pairs, and put it under a root sign, like so: \sqrt{01 09.00 00 00 00}

    Take the first pair of digits 01. Can we find a number d such that d^2 <= 01? Yes: d=1 fits the bill. Write the number 1 above the root.
    Below the root, underneath 01, write down 1 and subtract it giving 0. Now bring down the next pair of digits 09.
    Take the number above the line (1), double it to get 2.
    Can we find a units digit d such that d(2d) <= 09 (where 2d means "twenty plus d")? Well d has to be 0 because 1(21) = 21 > 09 (too big!)
    Write 0 above the line so we have 10 above the line. We've reached the decimal point after the 9 inside the root, so put a decimal point after the 10 above the line, so it says "10."

    Now subtract 0 from 09(!) to give 09 and bring down another pair of digits from under the root to give 900.
    Take the number above the line (10) and double it to get 20
    Can we find a units digit d such that d(20d) <= 900. Well 4(204) = 816 and 5(205) = 1025 so 5 is too big and we take the number 4 as d.
    Write 4 above the line so we have 10.4 above the line currently.

    Below the line, subtract 816 from 900 to give 84 and bring down another pair of digits to give 8400.
    Take the number above the line (ignoring the decimal point this is 104) and double it giving 208.
    Can we find a units digit d such that d(208d) <= 8400?
    Again 5 is too big, but 4(2084) = 8336 so write 4 above the line giving 10.44

    Below the line, subtract 8336 from 8400 leaving 64 and bring down another pair of digits to give 6400.
    Take the number above the line, without the decimal point i.e. 1044, and double it to give 2088.
    Can we find a units digit d such that d(2088d) <= 6400?
    In this case 1 is too big because 1(20881) = 20881 > 6400, so our next digit is 0. Write this above the line to give 10.440

    Below the line, subtract 0 from 6400(!) giving 6400 and bring down another pair of digits to give 640000.
    Take the number above the line (10440) and double it to get 20880.
    Can we find a units digit d such that d(20880d) <= 640000.
    In this case 4 is too big, so d = 3 because 3 x 208803 = 626409
    Write the 3 above the line to give: 10.4403

    So, to 3d.p. the answer is 10.440

    (quick check on calculator - answer is 10.4403065 approx)

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    (Original post by davros)
    You use long-square-rooting, the same way as you do long division in primary school.

    Write down your number with a leading zero and some extra zeros after the decimal point because we need to deal with the digits in pairs, and put it under a root sign, like so: \sqrt{01 09.00 00 00 00}

    Take the first pair of digits 01. Can we find a number d such that d^2 <= 01? Yes: d=1 fits the bill. Write the number 1 above the root.
    Below the root, underneath 01, write down 1 and subtract it giving 0. Now bring down the next pair of digits 09.
    Take the number above the line (1), double it to get 2.
    Can we find a units digit d such that d(2d) <= 09 (where 2d means "twenty plus d")? Well d has to be 0 because 1(21) = 21 > 09 (too big!)
    Write 0 above the line so we have 10 above the line. We've reached the decimal point after the 9 inside the root, so put a decimal point after the 10 above the line, so it says "10."

    Now subtract 0 from 09(!) to give 09 and bring down another pair of digits from under the root to give 900.
    Take the number above the line (10) and double it to get 20
    Can we find a units digit d such that d(20d) <= 900. Well 4(204) = 816 and 5(205) = 1025 so 5 is too big and we take the number 4 as d.
    Write 4 above the line so we have 10.4 above the line currently.

    Below the line, subtract 816 from 900 to give 84 and bring down another pair of digits to give 8400.
    Take the number above the line (ignoring the decimal point this is 104) and double it giving 208.
    Can we find a units digit d such that d(208d) <= 8400?
    Again 5 is too big, but 4(2084) = 8336 so write 4 above the line giving 10.44

    Below the line, subtract 8336 from 8400 leaving 64 and bring down another pair of digits to give 6400.
    Take the number above the line, without the decimal point i.e. 1044, and double it to give 2088.
    Can we find a units digit d such that d(2088d) <= 6400?
    In this case 1 is too big because 1(20881) = 20881 > 6400, so our next digit is 0. Write this above the line to give 10.440

    Below the line, subtract 0 from 6400(!) giving 6400 and bring down another pair of digits to give 640000.
    Take the number above the line (10440) and double it to get 20880.
    Can we find a units digit d such that d(20880d) <= 640000.
    In this case 4 is too big, so d = 3 because 3 x 208803 = 626409
    Write the 3 above the line to give: 10.4403

    So, to 3d.p. the answer is 10.440

    (quick check on calculator - answer is 10.4403065 approx)

    :eek:
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    (Original post by davros)
    You use long-square-rooting, the same way as you do long division in primary school.

    Write down your number with a leading zero and some extra zeros after the decimal point because we need to deal with the digits in pairs, and put it under a root sign, like so: \sqrt{01 09.00 00 00 00}

    Take the first pair of digits 01. Can we find a number d such that d^2 <= 01? Yes: d=1 fits the bill. Write the number 1 above the root.
    Below the root, underneath 01, write down 1 and subtract it giving 0. Now bring down the next pair of digits 09.
    Take the number above the line (1), double it to get 2.
    Can we find a units digit d such that d(2d) <= 09 (where 2d means "twenty plus d")? Well d has to be 0 because 1(21) = 21 > 09 (too big!)
    Write 0 above the line so we have 10 above the line. We've reached the decimal point after the 9 inside the root, so put a decimal point after the 10 above the line, so it says "10."

    Now subtract 0 from 09(!) to give 09 and bring down another pair of digits from under the root to give 900.
    Take the number above the line (10) and double it to get 20
    Can we find a units digit d such that d(20d) <= 900. Well 4(204) = 816 and 5(205) = 1025 so 5 is too big and we take the number 4 as d.
    Write 4 above the line so we have 10.4 above the line currently.

    Below the line, subtract 816 from 900 to give 84 and bring down another pair of digits to give 8400.
    Take the number above the line (ignoring the decimal point this is 104) and double it giving 208.
    Can we find a units digit d such that d(208d) <= 8400?
    Again 5 is too big, but 4(2084) = 8336 so write 4 above the line giving 10.44

    Below the line, subtract 8336 from 8400 leaving 64 and bring down another pair of digits to give 6400.
    Take the number above the line, without the decimal point i.e. 1044, and double it to give 2088.
    Can we find a units digit d such that d(2088d) <= 6400?
    In this case 1 is too big because 1(20881) = 20881 > 6400, so our next digit is 0. Write this above the line to give 10.440

    Below the line, subtract 0 from 6400(!) giving 6400 and bring down another pair of digits to give 640000.
    Take the number above the line (10440) and double it to get 20880.
    Can we find a units digit d such that d(20880d) <= 640000.
    In this case 4 is too big, so d = 3 because 3 x 208803 = 626409
    Write the 3 above the line to give: 10.4403

    So, to 3d.p. the answer is 10.440

    (quick check on calculator - answer is 10.4403065 approx)

    What is this dark magic?!!
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    (Original post by joostan)
    What is this dark magic?!!
    No, it's just the way people worked out square roots before calculators were invented - you just had to "man up" and get down to it.

    I think they should bring this back in primary schools at the same time as they do long division
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    (Original post by davros)
    No, it's just the way people worked out square roots before calculators were invented - you just had to "man up" and get down to it.

    I think they should bring this back in primary schools at the same time as they do long division
    Lol, I wish I'd learnt this in school, it'd be a nice party trick
    The way I used to do it is try and work out what multiplies to give the answer - trial and improvement. This method is much better
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    It is years since I last saw that method. Didn`t understand it then and still don`t.
    I would have written 109 as 100(1 + 0.09) and then used the binomial theorem.
 
 
 
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