Hello. In this lesson, I’ll be covering a few examples covering the solution of quadratic equations. Nothing fancy.

**Example 1** Solve the equation x^{2} = 4.

**Solution **Easy. I’ll first factorize it into two linear expressions, and then equate each factor to zero to get the roots.

The equation is equivalent to x^{2} – 4 = 0 or (x – 2)(x + 2) = 0. This gives us **x = 2** and **x = –2**.

I hope that you’ve realized that this factorization step isn’t required. We can directly solve the equation as follows:

x^{2} = 4 => **x = ****±2**

That is, every quadratic equation of the form x^{2} = a has the solution x = ± \(\sqrt{a}\). No need to factorize any more.

Now, I would like to bring to your attention a common misconception here.

What people do is this: x^{2} = 4 => x = \(\sqrt{4}\) (taking square root of both sides) => x = ±2, and later on concluding that \(\sqrt{4}\) = ±2.

This is incorrect. \(\sqrt{4}\) equals 2 and not ±2. The \(\sqrt{ }\) sign denotes the positive square root. So, what’s the correct way then?

x^{2} = 4 => x = ± \(\sqrt{4}\) => x = ± 2. The ± sign comes from the quadratic equation and not after ‘removing’ the square root.

I’ll explain again.

x^{2} – 4 = 0 => (x – \(\sqrt{4}\) )(x + \(\sqrt{4}\)) = 0 => x = \(\sqrt{4}\) or x = – \(\sqrt{4}\) . Combining these, we get x = ± \(\sqrt{4}\) or x = ±2.

In case you’re interested in taking the square root of both sides, we’ll get \(\sqrt{x^2}\) = \(\sqrt{4}\) => |x| = 2 => x = ±2. Remember that \(\sqrt{x^2}\) = |x|.

**Example 2 **Solve the equation x^{2} – 8x = 0.

**Solution **This one is easy too. Let’s factorize again.

The equation becomes x(x – 8) = 0, which gives **x = 0** and **x = 8**.

Now here’s another typical mistake people make: x^{2} – 8x = 0 implies x^{2} = 8x. And after ‘cancelling’ x from both sides, we get x = 8.

Well, this is wrong. Why? Because we lost a precious root (0) there – a quadratic equation is supposed to have two roots.

And what exactly did we do wrong? Cancelling an unknown term which could possibly have been zero.

Here’s the rule: You’re not allowed to cancel any term from both sides of an equation unless it is a non-zero term.

Otherwise, strange things will happen: 0 = 0 => 4 x 0 = 5 x 0 => 4 x ø = 5 x ø => 4 = 5. Very strange things.

To play safe, you should bring all the terms to one side, factorize, and equate all the factors to zero.

Let’s move on to the next example.

**Example 3 **Solve the equation x^{2} + 6x + 5 = 0.

**Solution **I’ll not use the quadratic formula yet. I’ll try to convert this equation to a form similar to the one in the first example.

Adding 9 to both sides gives me x^{2} + 6x + 9 + 5 = 9. This becomes (x + 3)^{2} + 5 = 9, or (x + 3)^{2} = 4.

Now you know what to do next, right?

We get x + 3 = ± 2. Or x = ±2 – 3. This gives **x = **–**1** and **x = **–**5**.

The method I used here is known as **completing the perfect square**.

That is, if you see something like a^{2} + 2ab, add and subtract b^{2} to get (a + b)^{2} – b^{2}, thereby completing the perfect square (a + b)^{2}.

Will this method always work? Yes.

And that’s the very idea behind the quadratic formula (in fact, any math formula). That is, make a formula out of a (sure shot) method’s end result, to save time.

And that is what I did in the previous lesson, when I added and subtracted ( \( \frac{b}{2a} \) )^{2}** – **completed the perfect square, found the roots, and preserved the formula.

We’re now officially qualified to use the quadratic formula.

**Example 4 **Solve the equation 2x^{2} + x – 1 = 0.

**Solution **If we compare this to the general form, i.e. ax^{2} + bx + c = 0, we have a = 2, b = 1 and c = **–**1.

Now I’ll just plug these values into the quadratic formula, i.e. \( \frac{-b\pm\sqrt{b^2-4ac}}{2a} \)

We get x = \( \frac{-1\pm\sqrt{1^2-4(2)(-1)}}{2(2)} \) = \( \frac{-1\pm3}{4} \) . This gives **x = –1** and **x = 1/2**. Pretty neat, right?

And that’ll be all for this lesson. In the next one, I’ll talk about equations that are convertible to quadratic equations. See you there.