Algebra Know-It-ALL

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

and

3 x− 2 = 0

The solution to the first of these equations is derived like this:

4 x+ 5 = 0

4 x=− 5

x=−5/4

In the second case, the process is similar:

3 x− 2 = 0

3 x= 2

x= 2/3

The roots of the quadratic are −5/4 and 2/3, and the solution set is {−5/4, 2/3}.

5. We want to morph the following quadratic so the left side becomes a product of a
   binomial with itself:

16 x^2 − 40 x+ 25 = 0

The coefficient of x^2 is equal to 16, and we know it has to be the square of the first term in

the binomial. That means the first term must be 4x or − 4 x. The constant in the polyno-

mial is equal to 25, and it must be the square of the constant in the binomial. That means

the constant in the binomial must be 5 or −5. The coefficient of x in the polynomial is

negative, telling us that the coefficient and the constant in the binomial must have oppo-

site signs. As things work out, we get

(4x− 5)^2 = 0

Checking to be sure this is right, we can multiply it out:

(4x− 5)(4x− 5) = 0

16 x^2 + (− 20 x)+ (− 20 x)+ 25 = 0

16 x^2 − 40 x+ 25 = 0

It works! We can also use

(− 4 x+ 5)^2 = 0