1001 Algebra Problems.PDF

661. c.Note that,x+ 21 = 10x

(^12) 
or equivalently,x–
10 x
(^12) 

• 21 = 0 can be written as (x
  (^12) 
  )^2 – 10(x
  (^12) 
  ) +
  21 = 0. Let u=x
  (^12) 
  .Then, rewriting the original
  equation yields the equation u^2 – 10u+ 21 = 0,
  which is quadratic. Factoring the left side results
  in the equivalent equation (u– 3)(u– 7) = 0.
  Solving this equation foruyields the solution
  u = 3 or u= 7. In order to solve the original
  equation, we go back to the substitution and
  write u in terms of the original variable x:
  u= 3 is the same as x
  (^12) 
  = 3, which gives x= 3^2 = 9
  u = 7 is the same as x
  21 
  = 7, which gives x= 7^2 = 49
  The solutions of the original equation are
  x= 9, 49.

662. a.Observe that 16 – 56x+ 49x= 0 can be
     written as 16 – 56x+ 49(x)^2 = 0. Let u=
     x. Rewriting the original equation yields the
     equation 16 – 56u+ 49u^2 = 0, which is quadratic.
     Factoring the left side results in the equivalent
     equation (4 – 7u)^2 = 0. Solving this equation
     foruyields the solution u= ^47 . In order to solve
     the original equation, we go back to the substi-
     tution and write u in terms of the original
     variable x:

u = ^47 is the same as x= ^47 , which gives us

x= (^47 )^2 = ^1469 

Therefore, solution of the original equation is

x= ^1469 .

663. a.Note that x– x= 6, or equivalently x–
     x– 6 = 0, can be written as (x)^2 – (x) –
     6 = 0. Let u = x.Then, rewriting the above
     equation yields the equation u^2 – u– 6 = 0,
     which is quadratic. Factoring yields the equiv-
     alent equation (u– 3)(u+ 2) = 0. Solving this
     equation for uyields the solutions u = –2 or
     u= 3. In order to solve the original equation,
     we must go back to the substitution and write
     u in terms of the original variable x:

u = –2 is the same as x= –2, which does not

have a solution

u = 3 is the same as x= 3, so that x= 9

Therefore, the solution of the original equa-

tion is x= 9.

664. c.We must first write the equation in the cor-
     rect form:

2 x

(^16) 

• x

(^13) 

• 1 = 0
  x

(^13) 

• 2x

(^16) 

• 1 = 0
  (x
  (^16) 
  )^2 – 2(x
  (^16) 
  ) + 1 = 0
  Next, let u = x
  (^16) 

. Then, we must solve the equa-
tion u^2 – 2u+ 1 = 0. Observe that factoring
this equation yields (u– 1)^2 = 0. Consequently,
u = 1. Next, we must go back to the actual sub-
stitution and solve the new equations obtained
by substituting in this value ofu. Specifically,
we must solve x

(^16) 
= 1. This is easily solved by
raising both sides to the power 6. The result is
that x= 1.
ANSWERS & EXPLANATIONS–