313. We have now to treat of the method of finding the Square Root of a Binomial Surd, that is, of an expression of one of the following forms:

m+ √n, m−√n,

where m stands for a whole or fractional number, and n for a surd of the second order.

314. We have first to prove two Theorems.

THEOREM I. If √a=m+√n, m must be zero.

that is, √n, a surd, is equal to a whole or fractional number, which is impossible.

Hence the assumed equality can never hold unless m= = 0, in which case √a=√n.

THEOREM II. If b+√a=m+√n, then must b=m, and √a=√n.

which, by Theorem I., is impossible unless x=0, in which case b=m and √a=√n.

a

315. To find the Square Root of a+√√b.

:. √(a+√b) = √ {a+√(a2 = b) } + √ { a = √(a2—b)}.

Similarly we may shew that

√(a−√b) = √√ {a+√(a2 − b ) } - √√ {a-√(a3-b)}.

316. The practical use of this method will be more clearly seen from the following example.

Find the square root of 18+2√(77).

√{18+2√(77)}=√∞+√y.

18+2√(77)=x+2√(xy)+y;

Assume

Then

That is, the square root required is √(11) +√7.

Find the square roots of the following Binomial Surds:

317. It is often easy to determine the square roots of expressions such as those given in the preceding set of Examples by inspection.

Take for instance the expression 18+2√(77).

What we want is to find two numbers whose sum is 18 and whose product is 77: these are evidently 11 and 7.

Then

18+2√(77)=11+7+2√(11x7)

={√(11) +√/7}3.

That is √(11)+√7 is the square root of 18+2√(77).

To effect this resolution by inspection it is necessary that the coefficient of the surd should be 2, and this we can always

ensure.

For example, if the proposed expression be 4+√(15), we proceed thus:

Again, to find the square root of 28–10√3.

28-10√3=28—2√(75)

=25+3-2√√(25 × 3)

=(5-√/3);

.. 5-√3 is the square root required.

318. ANY equation may be cleared of a single surd, by transposing all the other terms to the contrary side of the equation, and then raising each side to the power corresponding to the order of the surd.

The process will be explained by the following Examples. Ex. (1) √x=4.

Raising both sides to the second power,

13.

/(2x+3)+4=7.

14. b+c√x=a.

15. √(x-9)+x=9.

16. √(x2-11)=x−1.

✓ 17. √(4x2+5x-2)=2x+1.

18. √(9x2-12x−51)+3=3x.

19. √(x2—ax+b)—a=x.

20. √(25x2-3mx+n)−5x=m.

319. When two surds are involved in an equation, one at least may be made to disappear by disposing the terms in such a way that one of the surds stands by itself on one side of the equation, and then raising each side to the power corresponding to the order of the surd. If a surd be still left, it can be made to stand by itself, and removed by raising each side to a certain power.

Squaring both sides, x-5=36-12√√(x+7)+x+7

12√(x+7)=36+x+7−x+5,

12 √(x+7)=48,

therefore

or

or