Algebra Know-It-ALL

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

and

x+j 7 = 0

In the top equation, we can add j7 to each side, getting the root x=j7. In the bottom

equation, we can subtract j7 from each side, getting the root x=−j7. The roots can be

formally expressed this way:

x=j7 or x=−j 7

The solution set is

X= {j7,−j7}

2. To obtain the polynomial standard form of the equation stated in Prob. 1, we must
   multiply out the left side using the product of sums rule. We can minimize the risk
   of getting confused by the signs if we convert the first binomial factor to a sum. We
   proceed as follows:

[x+ (−j7)](x+j7)= 0

x^2 +xj 7 + (−j 7 x)+ (−j7)(j7) = 0

x^2 +j 7 x+ (−j 7 x)+ (−j×j)× 7 × 7 = 0

x^2 + 49 = 0

3. Remember the general polynomial standard form of a quadratic equation:

ax^2 +bx+c= 0

In the equation we derived in solution to Prob. 2, we have a= 1, b= 0, and c= 49.

Plugging these values into the quadratic formula and then working out the arithmetic,

we get

x= [−b± (b^2 − 4 ac)1/2] / (2a)

= [− 0 ± (0^2 − 4 × 1 × 49)1/2] / (2 × 1)

= [±(−196)1/2] / 2

=±j14/2

=±j 7

This agrees with the results we got when we solved Prob. 1.