Hello. In this lesson, I’ll be covering a few examples covering the solution of quadratic equations. Nothing fancy.
Example 1 Solve the equation x2 = 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 x2 – 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:
x2 = 4 => x = ±2
That is, every quadratic equation of the form x2 = 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: x2 = 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?
x2 = 4 => x = ± \(\sqrt{4}\) => x = ± 2. The ± sign comes from the quadratic equation and not after ‘removing’ the square root.
I’ll explain again.
x2 – 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 x2 – 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: x2 – 8x = 0 implies x2 = 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 x2 + 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 x2 + 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 a2 + 2ab, add and subtract b2 to get (a + b)2 – b2, 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 2x2 + x – 1 = 0.
Solution If we compare this to the general form, i.e. ax2 + 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.