1001 Algebra Problems.PDF

616. c.Square both sides of the equation and then
     solve for x:

10–3x= x–2

(10–3x)^2 = (x–2)^2

10–3x= x^2 –4x+ 4

0 = x^2 – x–6

0 =(x– 3)( x+ 2)

x= 3, –2

Substituting x = 3 into the original equation

yields the true statement 1 = 1, but substitut-

ing x = –2 into the original equation results in

the false statement 4 = –4. Only x= 3 is a solu-

tion to the original equation.

617. d.Square both sides of the equation and then
     solve for x:

 3 x+ 4+x= 8

 3 x+ 4= 8 – x

( 3 x+ 4)^2 = (8 – x)^2

3 x+ 4 = 64 – 16x+ x^2

0 = x^2 – 19x+ 60

0 = (x– 4)(x– 15)

x= 4, 15

Substituting x = 4 into the original equation

yields the true statement 8 = 8, but substitut-

ing x = 15 into the original equation results in

the false statement 22 = 8. Therefore, only x=

4 is a solution to the original equation.

618. b.Isolate the squared expression on one side
     and then, take the square root of both sides
     and solve for x:

(x– 1)^2 + 16 = 0

(x– 1)^2 = – 16

(x– 1)^2 = ±–16

x– 1 = ± 4i

x= 1 ± 4i

619. b.
     x^3 = – 27

(^3) x (^3) = ^3 –27
x= ^3 (–3)^3
x= – 3

620. c.
     x^2 = 225
     x^2 = ± 225 
     x= ± 15 


621. a.
     x^3 = –125
     ^3 x^3 = ^3 –125
     x= –5


622. c.
     (x+ 4)^2 = 81
     (x+ 4)^2 = ± 81 
     x+ 4 = ±9
     x= – 4 ±9
     x= 5, –13


623. d.
     x^2 + 1 = 0
     x^2 = – 1
     x^2 = ±– 1
     x= ±i^2
     x= ±i


624. b.
     x^2 + 81 = 0
     x^2 = –81
     x^2 = ±–81
     x= ±9i

ANSWERS & EXPLANATIONS–