Math 1010-2 Midterm 2A
Solution
Instructions: There are TEN problems in this exam. Answer the questions
in the spaces provided on the question sheets. If you need more space, use
the bottom of the last page, but remember to indicate which problem you are
doing. Partial credit will be awarded. Calculators are NOT allowed. This
exam is closed book and closed notes. Show you work on each problem.
1. (10 points) Solve the following equation:
−x2 + 3x − 3 = x
Solution: Minus by x on both sides of the equation
−x2 + 2x − 3 = 0
This is an standard quadratic equation, so we may apply quadratic formula to find the solutions.
√
−b ± b2 − 4ac
x=
2a
p
−2 ± 22 − 4(−1)(−3)
=
2(−1)
√
−2 ± −8
=
−2
√
−2 ± 2 2i
=
−2
√
= 1 ∓ 2i
So the two conjugate complex solutions are x = 1 +
√
2i or x = 1 −
2. (10 points) Simplify the following expression:
5x3 y 2
10(xy −1 )2
Solution:
5x3 y 2
5x3 y 2
=
10(xy −1 )2
10x2 y −2
5 x3 y 2
·
·
10 x2 y −2
1
= · x · y4
2
=
=
xy 4
2
3. (10 points) Simplify the following complex fraction:
x+3
x+5
1
x2 − 25
√
2i
Solution:
x+3
x + 3 x2 − 25
x+5
=
·
1
x+5
1
x2 − 25
(x + 3)(x2 − 25)
x+5
(x + 3)(x − 5)(x + 5)
=
x+5
=
= (x + 3)(x − 5)
= x2 − 2x − 15
4. (10 points) Rationalize the denominator of the following expression and simplify:
√
5+1
√
5−1
Solution:
√
√
√
( 5 + 1)( 5 + 1)
5+1
√
√
= √
5−1
( 5 − 1)( 5 + 1)
√
√
( 5)2 + 2 5 + 1
√
=
( 5)2 − 12
√
5+2 5+1
=
5−1
√
6+2 5
=
4
√
3+ 5
=
2
5. (10 points) Simplify the following product of polynomials by any method you choose:
(−x + x2 + 1)(x2 + x + 1)
Write your answer as a polynomial in standard form (i.e. from the highest power to the lowest).
Solution: Here we will use the vertical format. First, rewrite the two factors in standard form:
(x2 − x + 1)(x2 + x + 1).
x2 −x +1
×
x2 +x +1
x2 −x +1
3
x −x2 +x
x4 −x3 +x2
+x2
+1
x4
So the product is x4 + x2 + 1 .
Alternate Solution:
(−x + x2 + 1)(x2 + x + 1) = ((x2 + 1) − x)((x2 + 1) + x)
= (x2 + 1)2 − x2
= (x2 )2 + 2x2 + 1 − x2
= x4 + x2 + 1
6. (10 points) Factor the following polynomial:
x2 − 2x − 15
Solution: −15 can be separated into (−1) × 15, (−3) × 5, (−5) × 3 or (−15) × 1. We need to check
the following four possibilities:
(x − 1)(x + 15) = x2 + 14x − 15
(x − 3)(x + 5) = x2 + 2x − 15
(x − 5)(x + 3) = x2 − 2x − 15
(x − 15)(x + 1) = x2 − 14 − 15
×
×
√
×
So x2 − 2x − 15 = (x − 5)(x + 3) .
7. (10 points) Solve the following equation by any method you choose:
√
x− x−6=0
√
Solution: Since x = ( x)2 , the original equation can be rewritten as
√
√
( x)2 − x − 6 = 0
√
hence can be regarded as a standard quadratic equation in variable x. By factoring the trinomial,
we have
√
√
( x − 3)( x + 2) = 0
or
√
x−3=0
therefore
√
x=3
or
or
√
√
x+2=0
x = −2
take the square
x = 9 or
x=4
Check the solutions, only x = 9 is solution of the original equation.
Alternate Solution: We may solve it as a regular radical equation. First, isolate a radical:
√
x− x−6=0
√
x−6= x
√
(x − 6)2 = ( x)2
x2 − 12x + 36 = x
x2 − 13x + 36 = 0
(x − 4)(x − 9) = 0
So x = 4 or x = 9. Substitute to the original equation and find that only x = 9 is solution of the
original equation.
8. (10 points) Simplify the following radical expression:
√
√
3
3
5x4 + 40x
Solution:
√
3
5x4 +
√
3
√
√
3
3
40x = x3 · 5x + 23 · 5x
√
√
√
√
3
3
= x3 · 3 5x + 23 · 3 5x
√
√
= x · 3 5x + 2 · 3 5x
√
= (x + 2) 3 5x
9. (10 points) Evaluate the following expression so that no exponents appear in your final answer:
− 23
1
8
Solution:
− 23
1
1
= 2 =
8
1 3
8
1
1
1
r !2 = 2 = 1 = 4
1
3 1
4
2
8
10. (10 points) (Golden Section) A point separates a line segment into two parts as in the following
picture. Suppose the ratio between the length of the whole line segment and that of the left part is
the same as the ratio between the left part and the right part. And suppose the length of the right
part is 1. Find the length of the left part. (The ratio between a and b is ab .)
b
b
b
separation point
Solution: Suppose the length of the left part is x, then the length of the whole line segment will be
x + 1. As given in the problem,
left
whole
=
left
right
x+1
x
=
x
1
x + 1 = x2
x2 − x − 1 = 0
p
−(−1) ± (−1)2 − 4 · 1 · (−1)
x=
2·1
√
1± 5
x=
2
Since
√
5>
√
4 = 2 > 1, so
√
can only be
1+ 5
2
√
1− 5
2
< 0. But the length can not be a negative number, therefor it
. Search on Wikipedia for more information about Golden Section.