674 Worked-Out Solutions to Exercises: Chapters 21 to 29
- Let’s take the square root of both sides of the binomial factor equation from solution to
Prob. 7, remembering to include both the negative and positive results:
[(x+ 3)^2 ]1/2=±(161/2)
This simplifies to
x+ 3 =± 4
which can be stated as the following pair of first degree equations:
x+ 3 = 4 or x+ 3 =− 4
The first of these solves to x= 1, and the second solves to x=−7. The roots of the qua-
dratic are therefore x= 1 or x=−7, so the solution set is {1, −7}. Let’s check these roots
in the original quadratic. When x= 1, we get
x^2 + 6 x− 7 = 0
12 + 6 × 1 − 7 = 0
1 + 6 − 7 = 0
7 − 7 = 0
0 = 0
Whenx=−7, we get
x^2 + 6 x− 7 = 0
(−7)^2 + 6 × (−7)− 7 = 0
49 + (−42)− 7 = 0
49 − 42 − 7 = 0
7 − 7 = 0
0 = 0
- To determine how many real roots a quadratic has, we can calculate the discriminant.
For a quadratic of the form
ax^2 +bx+c= 0
the discriminant is b^2 − 4 ac. The equation of interest is
− 2 x^2 + 3 x+ 35 = 0
Here, a=−2,b= 3, and c= 35. Therefore
b^2 − 4 ac= 32 − 4 × (−2)× 35
= 9 − (−280)
= 9 + 280
= 289