Worked Examples

Today, we're going to solve the quadratic equation $x^2 + 5x + 6 = 0$ by factoring. Can anyone remind me of what factors are?

Factors are numbers we can multiply to get another number.

Exactly! Here, we need two numbers that multiply to 6 and add to 5. Can you think of any?

How about 2 and 3?

Great! So we can write the equation as $(x + 2)(x + 3) = 0$. What do we do next?

Set each factor to zero? So $x + 2 = 0$ and $x + 3 = 0$.

Exactly! Solving these gives us the roots. What are they?

So $x = -2$ and $x = -3$!

Excellent, everyone! Remember: when factoring, look for pairs of numbers that work. Now let's summarize: we factored the quadratic into binomials and solved for the roots!

Now let’s try the equation $x^2 + 4x + 1 = 0$ and solve it by completing the square. Who can recall what that means?

It’s about rearranging the equation to form a perfect square.

Correct! First, we move the constant to the other side: $x^2 + 4x = -1$. Now, to complete the square, we need to calculate half of 4 squared. What’s that?

That would be 4! So we add 4 to both sides.

Exactly! Now, we have $(x + 2)^2 = 3$. What’s our next move?

We take the square root of both sides to get $x + 2 = \, \pm \sqrt{3}$!

Fantastic! Now let’s isolate $x$ to find the solutions.

So $x = -2 \pm \sqrt{3}$!

Well done! Summarizing: we completed the square to find the roots.

Let’s tackle the equation $2x^2 - 4x - 6 = 0$ using the quadratic formula. Recall the formula is $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. Who can tell me the values of a, b, and c here?

Here, $a = 2$, $b = -4$, and $c = -6$.

Exactly! Now let’s substitute these values into our formula step-by-step. What does that look like?

So, we have $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(2)(-6)}}{2(2)}$.

Wonderful! Now simplify the discriminant first, what do you get?

That gives $x = \frac{4 \pm \sqrt{16 + 48}}{4} = \frac{4 \pm 8}{4}$.

Correct! So identifying the solutions, what do you find?

Then, we have $x = 3$ and $x = -1$!

Exactly right! So to summarize: we employed the quadratic formula and identified our roots.