# Elementary Algebra Exercise Book I by Bookboon.com

By Bookboon.com

Best elementary books

Notes on Rubik's Magic Cube

Notes on Rubik's 'Magic dice'

Beginner's Basque

This identify incorporates a ebook and a pair of audio CDs. Basque is the language spoken by way of the Basque those that reside within the Pyrenees in North valuable Spain and the adjacent area of south west France. it's also spoken through many immigrant groups around the globe together with the united states, Venezuela, Argentina, Mexico and Colombia.

Elementary Algebra

Trouble-free Algebra is a piece textual content that covers the normal themes studied in a contemporary uncomplicated algebra path. it truly is meant for college kids who (1) don't have any publicity to easy algebra, (2) have formerly had a nasty adventure with straight forward algebra, or (3) have to evaluate algebraic ideas and methods.

Additional resources for Elementary Algebra Exercise Book I

Sample text

14 Solve the equation 5x+1 = 3x Solution: 5x+1 = 3x Equations 2 −1 . 2 −1 ⇔ (x + 1) lg 5 = (x2 − 1) lg 3 ⇔ (x + 1)[lg 5 − (x − 1) lg 3] = 0 , thus x + 1 = 0 or lg 5 − (x − 1) lg 3 = 0 , which imply two solutions x = −1 , x = log3 15 . 15 Solve the equation x4 − 4x2 + 1 = 0 . Solution: Let y = x2 , then the equation becomes √ √ √ y 2 − 4y + 1 = 0 ⇔ (y − 2)2 = 3 ⇒ y = 2 ±√ 3 ⇒ x2 = 2 + 3 or x2 = 2 − 3 , √ of which implies the first of which implies x = ± 2 + 3 = ± 3+1 2 , the second √ √ √ √ √ √ 3+1 3+1 3−1 3−1 3−1 , − , , − .

Then x(x + 2) = z + 120 ⇔ z = (x − 10)(x + 12) . Since z is a prime number, then x − 10 = 1 , then x = 11, z = 23 . As a conclusion, there are two possibilities: x = 2, y = 59, z = 2 or x = 11, y = 2, z = 23 . 31 Solve the equation √ 2x2 − 7x + 1 − √ 2x2 − 7x + 1 + 2x2 − 9x + 4 : √ √ √ 2x2 − 7x + 1 + 2x2 − 9x + 4 = 2x − 3 (ii). (i)+(ii): 2x2 − 7x + 1 = x − 1 , taking square to obtain x2 − 5x = 0 ⇒ x(x − 5) = 0 ⇒ x = 0 or x = 5 . We can verify these two possible solutions via the original equation (i): x = 5 is indeed a root of (i), but x = 0 is a extraneous root Solution: Multiply both sides by √ √ 2x2 − 9x + 4 = 1 (i).

Since 2 and 5 are coprime, thus at least one of a3 b − ab3 = ab(a2 − b2 ), b3 c − bc3 = bc( = ab(a2 − b2 ), b3 c − bc3 = bc(b2 − c2 ), c3 a − ca3 = ca(c2 − a2 ) is divisible by 10. 108 Let a, b, c are positive integers and follow a geometric sequence, and b − a is a perfect square, log6 a + log6 b + log6 c = 6 , find the value of a + b + c . Solution: log6 a + log6 b + log6 c = 6 ⇒ log6 abc = 6 ⇒ abc = 66 . In addition, b2 = ac , then b = 62 = 36, ac = 362 . In order to make 36 − a a perfect square, a can only be 11,27,32,35, and a is a divisor of 362 , thus a = 27 , then c = 48 .