Page:Elementary algebra (1896).djvu/236

218 Rh Nore. Since every quantity has two square roots equal in magnitude but opposite in sign, strictly speaking we should have

the square root of 16 + 2 55 = \pm (11 + 5),

the square root of 16 — 255 = \pm(11 — 5).

However, it is usually sufficient to take the positive value of the square root, so that in assuming a —b = /x — y it is understood that x is greater than y. With this proviso it will be unnecessary in any numerical example to use the double sign at the stage of work corresponding to equation (3) of the last example.

261. When the binomial whose square root we are seeking consists of two quadratic surds, we proceed as explained in the following example.

Ex. Find the square root of 175 — 147.

Since 175 — 147 =7(25 — 21) = 7(5 — 21),

175 — 147 = 4 7 (5 -21). Vole But 5 - 21 = {7}{2} - {3}{2} 175 — 147 = 4 7 ( {7}{2} - {3}{2}).

262. To find the square root of a binomial surd by inspection.

Ex. 1. Find the square root of 11 + 2 30.

We have only to find two quantities whose sum is 11, and whose product is 30, thus

11+2 30=6+4+5+2 6x5=(6 +5)^2. 11+ 2 30 = 6 + 5.

Ex. 2. Find the square root of 53 — 12 10.

First write the binomial so that the surd part has a coefficient 2 ;

thus 53 — 12 10 = 53 — 2 360.

We have now to find two quantities whose sum is 53 and whose product is 360 ; these are 45 and 8;

hence 58 — 12 10 = 45 + 8 —2 45 8 =(45 — 8)^2; 53 — 12 10 = 45 — 8 = 3 5 — 2 2.