MATH 74 HOMEWORK 1: DUE WEDNESDAY 9/5
1(a).
Find all real numbers
x
satisfying
x
4

6
x
2
+ 7 = 0.
If
x
satisﬁes the given equation, then the real number
t
=
x
2
satisﬁes
t
2

6
t
+7 =
0. Applying the proposition and simplifying a little, we see that either
t
= 3 +
√
2
or
t
= 3

√
2; we conclude that either
x
2
= 3 +
√
2 or
x
2
= 3

√
2. Applying the
proposition again in either of these cases, or just taking square roots, we see that
x
must one of the numbers
p
3 +
√
2,

p
3 +
√
2,
p
3

√
2,

p
3

√
2.
Conversely, if
x
happens to be any one of those numbers listed, we see that
x
2
is one of the numbers 3
±
√
2, so that (by the proposition again)
x
4

6
x
2
+ 7 = 0.
Conclusion: the set of real numbers
x
satisfying the equation is precisely the set
{
p
3 +
√
2
,

p
3 +
√
2
,
p
3

√
2
,

p
3

√
2
}
.
1(b).
Find all real numbers
x
satisfying
x
4

2
x
2

1 = 0.
If
x
is a real number satisfying the given equation, then
t
=
x
2
satisﬁes
t
2

2
t

1 = 0, and hence (by the proposition and some algebra) is one of the numbers
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 Fall '07
 COURTNEY
 Real Numbers, Proposition, Complex number, real solutions, tht x2

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