```MATH 1010-2: Quiz 9A
Solution
1. (7 points) Solve the following equation using any method you choose.
x2 + 2x + 5 = 0
x=
−b ±
√
b2 − 4ac
2a
where a = 1, b = 2 and c = 5. Substitute the value of a, b and c into
√
−2 ± 22 − 4 · 1 · 5
x=
2·1
√
−2 ± −16
=
2
−2 ± 4i
=
2
= −1 ± 2i
In conclusion, the two solutions of the original equation are −1 + 2i and −1 − 2i.
2. (3 points) Solve the following equation using any method you choose.
√
x− x−2=0
√
Solution: Since x = ( x)2 , the original equation can be rewritten as
√
√
( x)2 − x − 2 = 0
hence can be regarded as a standard quadratic equation in variable
By factoring the trinomial, we have
√
√
( x − 2)( x + 1) = 0
or
therefore
√
x − 2 = 0 or
√
x=2
or
√
√
x+1=0
x = −1
take the square
x=4
or
x=1
Check the solutions, only x = 4 is solution of the original equation.
√
x.
MATH 1010-2: Quiz 9B
Solution
1. (7 points) Solve the following equation using any method you choose.
x2 − 2x + 5 = 0
x=
−b ±
√
b2 − 4ac
2a
where a = 1, b = −2 and c = 5. Substitute the value of a, b and c into
p
−(−2) ± (−2)2 − 4 · 1 · 5
x=
2·1
√
2 ± −16
=
2
2 ± 4i
=
2
= 1 ± 2i
In conclusion, the two solutions of the original equation are 1 + 2i and 1 − 2i.
2. (3 points) Solve the following equation using any method you choose.
√
x+ x−2=0
√
Solution: Since x = ( x)2 , the original equation can be rewritten as
√
√
( x)2 + x − 2 = 0
hence can be regarded as a standard quadratic equation in variable
By factoring the trinomial, we have
√
√
( x + 2)( x − 1) = 0
or
therefore
√
x+2=0
√
or
x = −2 or
√
x−1=0
√
x=1
take the square
x=4
or
x=1
Check the solutions, only x = 1 is solution of the original equation.
√
x.
```