Set 42 (Page 98)
- b.Observe that b^4 – 7b^2 + 12 = 0 can be written
as (b^2 )^2 – 7(b)^2 + 12 = 0. Let u= b^2. Then, rewrit-
ing the previous equation yields the equation
u^2 – 7u+ 12 = 0, which is quadratic. Factoring
the left side results in the equivalent equation
(u– 4)(u– 3) = 0. Solving this equation for u
yields the solutions u= 4 oru= 3. In order to
solve the original equation, we must go back
to the substitution and write u in terms of the
original variable b:
u = 4 is the same as b^2 = 4, which gives us
b= ±2u = 3 is the same as b^2 = 3, which gives us
b= ± 3 The solutions of the original equation are
b= ±2, ± 3 .- a.Let u = b^2. Observe that (3b^2 – 1)(1– 2b^2 ) = 0
can be written as (3u– 1)(1– 2u) = 0, which is
quadratic. Solving this equation for uyields
the solutions u = ^13 or u= ^12 . In order to solve
the original equation, we must go back to the
substitution and write u in terms of the origi-
nal variableb:
u= ^13 is the same as b^2 = ^13 , which gives usb= ±^ = ±
u= ^12 is the same as b^2 = ^12 , which gives usb= ±^ = ±
The solutions of the original equation are
b= ±,±.- d.Note that 4b^4 + 20b^2 +25 = 0 can be written
as 4(b^2 )^2 + 20(b)^2 + 25 = 0. Let u= b^2 .Rewrit-
ing the original equation yields the equation,
4 u^2 + 20u+25 = 0, which is quadratic. Factor-
ing the left side results in the equivalent equa-
tion (2u + 5)^2 = 0. Solving this equation foru
yields the solution u = –^52 . Solving the original
equation requires that we go back to the sub-
stitution and write u in terms of the original
variable b:
u= –^52 is the same as b^2 = –^52 , which gives usb= ± – = ± i= ±i()
Therefore, the solutions of the original equa-
tion are b= ±i().- b.Observe that 16b^4 – 1 = 0 can be written as
16(b^2 )^2 – 1 = 0. Let u = b^2. Rewriting the origi-
nal equation yields the equation 16u^2 – 1 = 0,
which is quadratic. Factoring the left side results
in the equivalent equation (4u– 1)(4u+ 1) = 0.
Solving this equation for uyields the solution
u= ±^14 . In order to solve the original equation,
we must go back to the substitution and write
u in terms of the original variable b:
u= –^14 is the same as b^2 = –^14 , which gives usb= ± – = ±i ±i(^12 )
u= ^14 is the same as b^2 = ^14 , which gives usb= ± = ±^12
The solutions of the original equation are
b= ±i(^12 ), ±^12 .^1
4^1
4
^1
4^10
2^10
2
^5
2
^5
2^3
3
^2
2^2
2
^1
2^3
3
^1
3ANSWERS & EXPLANATIONS–