GMAT Quantitative Section Урок 50 Arithmetic square root
Posted in GMAT by Березень 28th, 2008

GMAT Quantitative Section Урок 50 Arithmetic square root В деяких завданнях на тесті GMAT в математичній частині ви зіткнетеся з необхідністю роботи з арифметичним квадратним коренем . Для успішного вирішення таких завдань потрібно володіти певними навиками. Саме їм і присвячений сьогоднішній урок.
Давайте розглянемо теорітечисекую основу даного поняття, підкріпивши практичними завданнями.


Today we will have а look at the most common tasks that contain arithmetic square root and learn, how to solve them in the shortest possible period of time.

(символ V позначає квадратний корінь – прим. редактора)


At first let’s look at the most simple tasks, in which you have only to count square root of some expression.

V(16*20 + 8*32)= ?

(A) 4V20

(B) 22 24

(C) 25

(D) 4V20 + 8V2

(E) 32

Since 16*20 + 8*32 = 576 = 242, V(16*20 + 8*32)= 24. The answer is B.

The second variant of solving this problem is to split off expression inside V: 16*20 + 8*32 = 16*(20 + 16)= 16*36 = 42*62 = (4*6)2, the answer is the same: 24.


Sometimes it is useful to remember algebraic expressions and formulas and use them to solve problems with roots.

Following facts should be useful:



a2 – b2 = (а + b)(а – b); (а + b)2 = a2 + 2ab + b2; (ab)2 = a2b2; а(b + з)= ab + ас.



Sample problems:

(1 – V5)(1 + V5) = ?

(A) -4

(B) 2

(C) 6

(D) -4 – 2V5

(E) 6 – 2V5

Obviously (1 – V5)(1 + V5) = 12 – (V5) 2 =1 – 5 = -4. A is the answer.


V(24+5V23)+V(24-5V23)=?

(A) 48

(B) V24

(C) 1

(D) 5V2

(E) 24-25V23

In order to rid the expression of square roots, let’s first square the entire expression. We are allowed to do this as long as we remember to “unsquare” whatever solution we get at that end.

V(24+5V23)+V(24-5V23) -> (V(24+5V23)+V(24-5V23))2


Notice that the new expression is of the form (x+y) 2 where x=V(24+5V23) and y=V(24-5V23).

Recall that (x+y) 2=x2+y2+2xy. This is one of the GMAT’s favorite expressions.

Returning to our expression:

x2=24+5V23, while y2=24-5V23 and 2xy=2(24+5V23)(24-5V23).

Notice that x2+y2 neatly simplifies to 48. This leaves only the 2xy expression left to simplify.

In order to simplify 2(24+5V23)(24-5V23), recall that (Va)(Vb)=V(ab).

Thus, 2(24+5V23)(24-5V23)=2(V((24+5V23)*(24-5V23)).

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