Can you give an example of a problem using either completing the square or the quadratic formula?

Question

Can you give an example of a problem using either completing the square or the quadratic formula?

Answer 1

Let me give you several examples:

Find x when x² – 8x – 20 = 0.

x² – 8x = 20.  (transposing the 20)

x² – 8x + 16 = 36.  (adding 16 to both sides)

(x – 4)² = 36. (factorising the square of a binomial)

x – 4 = ±6. (taking the square-root of both sides)

x = 4 ± 6 (transposing the – 4)

= 10 or – 2. (combining terms)

Now, the question arising oftenest at this point: How did you know to add the 16 to both sides? The reason is that x² – 8x + 16 is a perfect square. It is, in fact, the only quadratic expression that begins with x² – 8x  and is also a perfect square.

To see this, look for a moment at these other examples of perfect squares:

x² + 2x + 1 = (x + 1)².

x² – 2x + 1 = (x – 1)².

x² + 4x + 4 = (x + 2)².

x² – 4x + 4 = (x – 2)².

x² + 6x + 9 = (x + 3)².

x² – 6x + 9 = (x – 3)².

Two characteristics that these all have in common is,

1) that they all begin with x², which is itself a perfect square; and,

2) that the co-efficient of x, in the middle term, is twice the square-root of the constant in the third term.

To re-state the second characteristic, above: the third term is the square of half the co-efficient of x, contained in the middle term.

If you reflect on it, you will see that these two characteristics hold for all quadratics that are squares of (x + k), where k is some constant, and that is the trick you need to know, to solve quadratics by the method of completing the square.

So, then: to make x² – 8x  into a perfect square, you add the square of half of 8; namely, the square of 4, which is 16.

Now, let’s do a harder one:

Find x, when 2x² – 9x = – 10.

Now, the first thing we see is that begins with 2x², not x².

One way to deal with this is to divide through by 2, and, then, solve in the usual way:

x² – (9/2)x = – 5.

Now, take the square of half of 9/2: that is,

(9/4)² = 81/16.

Add 81/16 to both sides:

x² – (9/2)x + 81/16= 1/16.

This easily factorises:

(x – 9/4)² = 1/16.

Taking square-roots, we get:

x – 9/4 = ±1/4.

Simplifying, x = 10/4 or 8/4

= 2 or 2½.

Now, a third example, to illustrate the power of the method of completing the square, as it pertains to complex or irrational roots:

Solve x² + x + 1 = 0.

x² + x = –1

x² + x + ¼ = –3/4  

x + ½ = ±√(–3/4) = ± ½ i√3

x = –½  Ã‚± ½ i√3.

Factorisation works easily, when there are integer roots, but, when they are not, many prefer either completing the square or, else, the quadratic formula.

Speaking of the quadratic formula, let’s consider the general case,

ax² + bx + c = 0.

Let’s see what we get, if we solve this by completing the square:

ax² + bx = –c.

x² + (b/a)x = –c/a.

x² + (b/a)x + (b/2a)² = (b/2a)² – c/a

= b²/4a² – c/a

= (b² – 4ac) / 4a².

(x + b/2a)² = (b² – 4ac) / 4a².

x + b/2a = ±√(b² – 4ac) / 2a.

x = [–b ±√(b² – 4ac) ] / 2a.

The result is the familiar quadratic formula. Let’s now use this formula.

Solve x² + x + 1 = 0.

This equals

ax² + bx + c, where a = 1, b = 1, and c = 1.

Then, according to formula,

x = [–1 ±√(1² – 4) ] / 2

= [–1 ±√(–3) ] / 2

= –½ ± ½ i√3,

which is the answer we got before.

Some will prefer to memorise the quadratic formula and use it whenever factorisation produces no easy result. Others, myself included, will either factorise or use the method of completing the square. As you can see, none of these methods is terribly difficult. Because preferences differ, and because I believe that comprehension is aided by considering several alternative methods, I prefer to present it all three ways and let the student choose what he likes best.

I hope that you will find these examples of help.