Solve the roots of the equation
= y ² – 5200
x73 ² = 5329, let = 73 for a trial value
xAt = 73 x = 73 ² – 5200 = 129
yThe slope of the equation = y ² – 5200 at any station :
xd/yd
= xd( ² – 5200)/xd
= 2xxAt = 73 , Slope = 2 × 73 = 146
xEquation of tangent at P ( , x ) = ( 73 , 129 ) :
y146 = ( – 129 ) / ( y – 73 )
x146 – 10658 = x – 129
y = 146y – 10529
xWhen = 0 the y-intercept of the tangent is :
x = 10529/146 = 72.11644
xTesting the result : 72.11644 ² = 5200.78092 Let 72.11644 be the new trial value At = 72.11644 x = 72.11644 ² – 5200 = .78092
yAt = 72.11644 , Slope = 2 × 72.11644 = 144.23288
xEquation of tangent at P ( , x ) = ( 72.11644 , .78092 ) :
y144.23288 = ( – .78092 ) / ( y – 72.11644 )
x144.23288 – 10401.56184 = x – .78092
y = 144.23288y – 10400.78092
xWhen = 0 the y-intercept of the tangent is :
x = 10400.78092/144.23288 = 72.11103
xTesting the result : 72.11103 ² = 5200.00065 |

Find the square root of 5200
The closest square to 5200 is 72 × 72 = 5184 5200 = 5184 + 16 5200 ^{½}
= ( 5184 + 16 )^{½}
= 5184^{½}
× ( 1 + 16/5184 )^{½}
= 72 × ( 1 + 16/5184 ) ^{½}
The terms in the parentheses are in the correct form, with = 16/5184 and n = 1/2 :
x72 × ( 1 + 16/5184 ) ^{½}
= 72 × ( 1 + 1/2 × 16/5184 + [ 1/2 × -1/2 × 16/5184 × 16/5184 ] / 2 + [ 1/2 × -1/2 × -3/2 × 16/5184 × 16/5184 × 16/5184 ] / [ 3 × 2 ] + ... ) = 72 × ( 1 + 1/648 – 1/839808 + 1/544195584 – ... ) = 72.11103 ... checking the result : 72.11103 ² = 5200.00065 Since = 16/5148 = 1/324 is small the series converges rapidly.
Two terms would still return the same accuracy to five decimal places and :
x72 × ( 1 + 1/648) = 72.11189 ... accurate to two decimal places |