Algebra Know-It-ALL

674 Worked-Out Solutions to Exercises: Chapters 21 to 29

8. 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

9. 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