1001 Algebra Problems.PDF

668. b.Let u=x^2 +x. Observe that (x^2 +x)^2 + 12 =
     8(x^2 +x)or equivalently (x^2 +x)^2 – 8(x^2 +x) +
     12 = 0, can be written as u^2 – 8u + 12 = 0,
     which is quadratic. Factoring yields the equiv-
     alent equation (u – 6)(u – 2) = 0. Solving this
     equation for uyields the solutions u = 2 or u=


669. To solve the original equation, we go back to
     the substitution and write u in terms of the
     original variable x. Doing so yields two more
     quadratic equations, this time in x, that must
     be solved: First,u = 2 is the same as x^2 +x= 2,
     or equivalently x^2 +x– 2 = 0. Factoring yields
     the equation (x+ 2)(x– 1) = 0, so that x= – 2
     or 1. Similarly,u= 6 is the same as x^2 + x= 6,
     or equivalently x^2 + x– 6 = 0. Factoring yields
     the equation (x+ 3)(x– 2) = 0, so that x= –3
     or 2. Therefore, the solutions of the original
     equation are x= –3, –2, 1, or 2.


670. a.Let u = 1 + w. Observe that 21 + w

2

= 131 + w– 6, or equivalently

(^2) 1 + w
2

• 131 + w+ 6 = 0, can be

written as 2u^2 – 13u+ 6 = 0, which is qua-

dratic. Factoring yields the equivalent equa-

tion (2u– 1)(u– 6) = 0. Solving this equation

for uyields the solutions u= ^12 or u= 6. Solving

the original equation requires that we go back

to the substitution and write u in terms of the

original variablew. Doing so yields two more

radical equations, this time in w, that must be

solved. First,u= ^12 is the same as 1 + w= ^12 .

Isolating the radical term yields w= –^12 ,

which has no solution. Similarly,u = 6 is the

same as 1 + w= 6. Isolating the radical term

yields w= 5, so that w= 25. The solution of

the original equation is w= 25.

670. a.Let u = r– ^3 r.Observe that (r– ^3 r)^2 – (r– ^3 r) –
     6 = 0 can be written as u^2 – u– 6 = 0, which is
     quadratic. Factoring yields the equivalent
     equation (u – 3)(u + 2) = 0. Solving this equa-
     tion for uyields the solutions u= – 2 oru= 3.
     In order to solve the original equation, we
     must go back to the substitution and write u in
     terms of the original variabler. Doing so yields
     two more equations involving rational expres-
     sions, this time in r, that must be solved. First,
     u= –2 is the same as r– ^3 r= –2. Multiply both
     sides by rand solve for r:

r– ^3 r= – 2

r^2 – 3 = – 2r

r^2 + 2r– 3 = 0

(r+ 3)(r– 1) = 0

r = –3, 1

Similarly,u= 3 is the same as r– ^3 r= 3. Multi-

ply both sides by rand solve forr:

r– ^3 r= 3

r^2 – 3 = 3r

r^2 – 3 r – 3 = 0

Using the quadratic formula then yields r =

=.

The solutions of the original equation are

r = –3, 1,3 ± 221 .

3 ±^21 

2

–(–3) ±(–3)^2 – 4(1)(–3)

2(1)

ANSWERS & EXPLANATIONS–