A quadratic equation is an equation of a second degree, meaning that the equation contains at least one unknown variable that is squared. Every quadratic equation generally can be written as

$$ ax^2+bx+c=0 $$

Where \(a\), \(b\) and \(c\) are constants and \(a\ne0\).

In this lesson we will firstly cover some methods for solving special cases of quadratic equations when \(b=0\) and/or \(c=0\). These are useful even when we know how to solve the general equation, because they require less time than using the general formula. When we know how to solve the special cases we will learn how to solve the general equation as well. Lastly, we will go through some equations which can be turned into quadratic equations, look at complex roots and learn how to use all this knowledge when solving different types of problems.

Let’s examine the following equation:

$$ x^2 = 9 $$

A certain number multiplied by itself becomes \(9\). One solution is \(x=\sqrt{9}\) or \(x=3\), and since a negative number multiplied by itself gives a positive product our second solution is \(x=(-3)\).

Whenever we take the square root out of both sides of the equation, it’s very important that we include both the positive and the negative root. Had we stated our answer as just

$$x^2=9$$

$$\sqrt{x^2}=\sqrt{9}$$

$$x=3$$

we would have missed the second solution. The right way to solve this equation is to include a plus-minus sign when taking the square root out of both sides:

$$x^2=9$$

$$\sqrt{x^2}=\pm \sqrt{9}$$

$$x=\pm 3$$

In our example \(x^2=9\), we did not write the equation in the general form shown in the introduction. It can however be written in that form as well.

$$ x^2=9 \Rightarrow x^2-9=0 $$

As you see here \(a=1\), \(b=0\) and \(c=(-9)\). If a quadratic equation looks like this, with \(b=0\), then you can solve it in a similar way as we did with \(x^2=9\).

$$ ax^2+c=0 $$

$$ ax^2=-c $$

$$ x^2= - \frac {c}{a} $$

$$ x= \pm \sqrt {- \frac {c}{a}} $$

When \(c=0\) then one root is zero. This is because x can be factored out, making the equation:

$$ ax^2+bx=0 \Rightarrow x(ax+b)=0 $$

The left-hand side of the equation will be equal to zero if and only if \(x=0\) or \(ax+b=0\). The two solutions are thus \(x_1 = 0\) and \(x_2 = - \frac{b}{a} \).

What we have learned so far is enough to solve quadratic equations with \(b=0\) or \(c=0\). In complete quadratic equations this is not possible.

We will start with two special cases of quadratic equations which can be factored with the conjugate rule and the binomial square identities.

The conjugate rule can be used when the equation is of the form \(x^2-a^2=0\). This factors as \((x+a)(x-a)=0\).

The binomial square identities can be used when the expression is a perfect square. \(x^2+2ax+a^2=({x+a})^2\) and \(x^2-2ax+a^2=({x-a})^2\).

Here are some exercises to practise on:

Now when we know how to use these rules, here are some new quadratic equations that we are able to solve:

What if the equation cannot be turned to factors by any of the ways we have learned so far? For example the equation \( x^2-2x-15=0 \).

Quadratic equations can generally be written as \(ax^2+bx+c=0\). But we can also write them in the factored form \(k(x-p)(x-q)=0\) where \(k\), \(p\) and \(q\) are constants.

Using the zero product property, we can easily see that the solutions to the equation are \(p\) and \(q\) because then one of the factors equals \(0\) (\(k\) cannot be \(0\) because then the whole expression would be \(0\)). If we have a quadratic equation which is either written in factored form, or if you can change it to factored form, you can easily solve it. In the previous equation, we learned how to factorise simpler quadratic expressions. Now we will learn how to factor more difficult expressions. This is useful because factoring is often the fastest technique for solving quadratic equations. To start with we will only look at equations on the form \(x^2+bx+c=0\) and \((x-p)(x-q)\) where there is no coefficient in front of the \(x^2\) term. Once we have mastered this factoring these equations, we will move on to equations that includes such a coefficient.

We factor \(x^2-2x-15\) as follows:

We have an expression written as \(x^2+bx+c\) and we want it in the form of \((x+p)(x+q)\). If we expand this expression we get:

$$ (x+p)(x+q)=x^2+px+qx+pq=x^2+(p+q)x+pq $$

Compare this to the general equation \(ax^2 + bx + c = 0\). As you can see, \(b=(p+q)\) and \(c=pq\).

Or in our case: \((p+q)=-2\) and \(pq=-15\)

We are thus looking for two numbers \(p\) and \(q\) where the sum is \( -2 \) and the product is \(-15\).

Often the best way to start is by factoring \(c\).

$$ -15=3 \cdot 5 \cdot (-1) $$

This gives us a clue that \(p\) and \(q\) are \( \pm 5\) and \( \pm 3\) where one is positive and the other is negative. Let us look at the other equation: \(p+q=-2\). And because \(5\) is \(2\) greater than \(3\) and \(p+q\) is a negative number, \(5\) should be negative. \(3+(-5)=3-5=-2\). \(x^2-2x-15\) can be factored as \((x-5)(x+3)\). The solutions are thus \(x_1=5\) and \(x_2=(-3)\).

When factoring a quadratic expression, what we strive to do is to split up the coefficient \(b\) into two numbers \(p\) and \(q\) so that the sum \(p+q\) is equal to \(b\) and the product \(pq\) is equal to the constant \(c\). In the example above, we notice that \(3+(-5)=-2\) and \(3 \cdot (-5) =-15\). From this we decide to split up \(b=-2\) in the original equations as:

$$x^2-2x-15$$

$$x^2+3x-5x-15$$

Now the terms \(x^2\) and \(3x\) share the common factor \(x\). Similarly, \(5x\) and \(-15\) share the common factor \(5\). Factoring this out gives us:

$$x(x+3)-5(x+3)$$

Now both terms \(x(x+3)\) and \(5(x+3)\) share the common factor \( (x+3) \) so we can factor this as:

$$(x+3)(x-5)$$

and we are done.

Not every equation is written in the form of \(x^2+bx+c=0\). The general form of quadratic equations is:

$$ ax^2 + bx + c = 0 $$

This can be solved similarly as before by factoring but now into the form of \((rx+p)(sx+q)\). If we expand this expression we end up with \(rsx^2+(rq+sp)x+pq\), so \(a=rs\), \(b=(rq+sp)\) and \(c=pq\).

It is a similar process as before, but with some differences. Let me show you with an example:

$$ 4x^2+8x+3=0 $$

In some cases it can be hard to factor with the technique above. In that case we can do a thing that is called completing the square. If we just look at the first two terms, \(x^2\) and \(-2x\), does that bring the binomial square identity to your mind? In that case, you are completely right. If it did not, do not worry, for this is exactly what you will learn here.

\(x^2-2x\)…. looks exactly like the beginning of \(x^2-2x+1\) which comes if you expand \({x-1}^2\). But the third term is different. In our example it is \(-15\) instead of \(1\). But what we have can be written as follows:

$$ x^2-2x-15 = x^2-2x+1-1-15 = $$$$ = x^2-2x+1-16 = ({x-1})^2-16 $$

And \(16 = 4^2\), so

$$ ({x-1})^2-16 = ({x-1})^2 – 4^2 $$

And what have we got here?

The conjugate rule.

$$ ({x-1})^2-4^2 = (x-1+4)(x-1-4) = (x+3)(x-5) $$

So \(x^2-2x-15 = (x+3)(x-5)\) and we have our factors which we can just put into the equation.

$$ x^2-2x-15 = 0 $$

$$ (x+3)(x-5) = 0 $$

\(x+3=0 \Rightarrow x=-3\)

\(x-5=0 \Rightarrow x=5\)

And we have got our answers.

You could solve quadratic equations by factoring but there is a formula called the abc formula which you could use instead. It is found by completing the square, so it is practically the same thing, only that you put in the numbers in a formula instead which often makes it less messy and thereby reduces the risk of mistakes.

$$ ax^2+bx+c=0 $$

$$ x^2 + \frac{b}{a} x + \frac{c}{a} = 0 $$

$$ x^2 + 2 \frac{b}{2a} x + \frac{c}{a} = 0 $$

$$ (x^2 + 2 \frac{b}{2a} x + {(\frac{b}{2a})}^2) - {(\frac{b}{2a})}^2 + \frac{c}{a} = 0 $$

$$ (x^2 + 2 \frac{b}{2a} x + {(\frac{b}{2a})}^2) = {(\frac{b}{2a})}^2 - \frac{c}{a} $$

$$ {(x + \frac{b}{2a})}^2 = {(\frac{b}{2a})}^2 - \frac{c}{a} $$

$$ x + \frac{b}{2a} = \pm \sqrt{{(\frac{b}{2a})}^2 - \frac{c}{a}} $$

$$ x = - \frac{b}{2a} \pm \sqrt{{(\frac{b}{2a})}^2 - \frac{c}{a}} $$

$$ x = - \frac{b}{2a} \pm \sqrt{\frac{b^2}{4a^2} - \frac{c}{a}} $$

$$ x = - \frac{b}{2a} \pm \sqrt{\frac{b^2}{4a^2} - \frac{4ac}{4a^2}} $$

$$ x = - \frac{b}{2a} \pm \sqrt{\frac{b^2-4ac}{4a^2}} $$

$$ x = - \frac{b}{2a} \pm \frac {\sqrt{b^2-4ac}}{2a} $$

And that is the abc-formula. With that, you can solve all quadratic equations.

Now you can solve complete quadratic equations. But we are not done yet. Now the fun begins.

I will now go back to the first example of the lesson about quadratic equations: \(x^2=9\). Let us now say instead that \(x^2=(-9)\). This equation doesn’t have any real solutions. There are however solutions, they are just not real. Real in the world of mathematics does not mean exactly the same as in our ordinary world, even though there are similarities. One can say that the real numbers can be found on the number line. The real numbers consist of the natural numbers (\(0\), \(1\), \(2\), \(3\)...), the rest of the integers (\(-1\), \(-2\), \(-3\)), the rest of the rational numbers (\(-\frac{2}{3}\) or \(\frac{31}{5}\)) and the irrational numbers (\(\pi\), \(e\)). They do not include the imaginary and the complex numbers, which are what we will talk about now. We need imaginary numbers when solving equations when we need to take the square root of a negative number. To be able to do this, mathematicians have invented a special number, an imaginary number, called \(i\).

Shortly, \(i=\sqrt{-1}\).

Therefore \(i\) is not a real number, you cannot put \(i\) on the number axis. You cannot have \(i\) apples. It cannot be \(i\) degrees warm. \(i\) is an imaginary number. One can say that it only exists in our imagination. But not really, because \(i\) can be very useful too. If you take \(i^2\) then you get \(-1\) which is a real number. So often \(i\) works as a middle hand between the real numbers. To make use of the imaginary numbers, we have to get to know them and learn how to count with them. In this lesson, which is about quadratic equations, I will not go through it all because there is another lesson about them and you do not need to know all about them to (almost) master quadratic equations. Some things you will learn here, however.

All imaginary numbers can be written as \(bi\) (\(b \cdot i\)). If the number also consists of a real part, if it is written as \(a+bi\), then it is called a complex number. Some quadratic equations have imaginary or complex solutions.

So back to \(x^2=(-9)\).

$$ (-9)=(-1) \cdot 9 $$

Now we have \(x^2=(-1) \cdot 9\)

$$ x= \pm \sqrt {(-1) \cdot 9} = \pm \sqrt{-1} \cdot \sqrt{9} $$

And from the definition, \(i=\sqrt{-1} \)

And \( \sqrt{9}=3 \)

So \(x=\pm (i \cdot 3) = \pm 3i \)

And now we have our two solutions: \(x = \pm 3i \)

It works almost the same when using the abc formula.

$$ x^2-2x+5=0 $$

But why are there not any real solutions? To show this we can draw two equations as functions in a coordinate system. The first with two real solutions and the second with two complex solutions.

\(y = x^2-2x-15\)

\(y = x^2-2x+5\)

We have our solutions when the curves intersect the x-axis, i.e. when \(y=0\).

The first curve intersects the x-axis where \(x=5\) and where \(x=-3\).

The second curve is always above the x-axis, it never intersects. Therefore there are not any real solutions to that equation because they can’t be found on the number line.

And when there is only one answer, for example in the equation \(x^2-2x+1=0\), then the function \(y=x^2-2x+1\) looks like this.

It just touches the x-axis at one point, where \(x=1\). Therefore that equation has only one unique solution, but it is a “double-root” and is counted as two solutions. Therefore every quadratic equation has two solutions, but they are not always real or distinct from one another.

The hunt for new equations with higher exponents does not stop at quadratic equations. There are equations with exponents greater than \(2\). Normally, those cannot be solved with the techniques for solving quadratic equations. But there are some special cases which can be solved in this way. An equation which can be solved using what we know now is

$$ x^4-2x^2-15=0 $$

This is the same equation I used earlier in the lesson, only that \(x^2\) is changed to \(x^4\) and \(x\) is changed to \(x^2\). In both cases, the exponent has doubled. That fact we can use. If we substitute \(x^2\) for \(y\) then we get

$$ x^4-2x^2-15={x^2}^2-2{x^2}-15=y^2-2y-15=0 $$

which is our old equation. And by solving this we get two possible \(y\) which solve the equation,

\(y = 5\)

\(y = (-3)\)

But we are not done yet. The equation we had from the start had the highest exponent \(4\), and therefore there are \(4\) solutions to find. And furthermore, we had \(x\) in the beginning but now we have \(y\). To get \(x\) we need to “un-do” what we did before. Before we used \(x^2=y\). Now we have to solve two new equations, the \(x^2=y\) with our two \(y\).

\( x^2=5 \Rightarrow x= \pm \sqrt{5} \)

\( x^2=(-3) \Rightarrow x= \pm \sqrt{-3} = \pm \sqrt{3 \cdot (-1)} \Rightarrow x= \pm \sqrt{3} \cdot i \)

We have four solutions, \(x=\sqrt{5}\), \(x=-\sqrt{5}\), \(x=\sqrt{3}\cdot i\), \(x=-\sqrt{3} \cdot i\).

Which of these equations can be solved by substituting \(x^2\) for \(y\)?

\( x^4+5x^3+1=0 \)

\( x^4+5x^2+1=0 \)

\( x^4+5x+1=0 \)

This is almost the same concept as above, only that it is not about solving special cases of \(x^4\)-equations but a way of solving certain quadratic equations, and some special equations faster and easier.

In the following exercises, solve by substituting a certain expression with \(x\) to \(y\).

As we saw in a previous part, equations can also be written as functions. In that case the \(0\) changes to \(y\) or \(f(x)\) and the solutions of the equation are the \(x\) where \(f(x)=0\). These are also called roots. Linear functions can be written as \(y=kx+m\) where \(k\) and \(m\) are constants. In the same way, quadratic equations can be written as \(y=ax^2+bx+c\) where \(a\), \(b\) and \(c\) are constants and \(a \ne 0\). Another way to write quadratic equations is in the form \(y=a{(x-h)}^2+k\). This can be found by completing the square. It is called vertex form because it is helpful when determining the vertex, or the extreme point. It is also helpful when finding symmetry. Quadratic equations are symmetrical, which (in this case) means that there can be drawn a vertical line and the left side is a reflection of the right side. To show how the vertex and the symmetry line can be found, here is an example:

Find the vertex and the symmetry line to the function \(y=a{(x-h)}^2+k\).

But why does this symmetry exist? The reason is that if \(x\) becomes one unit greater or one unit less than \(h\), the value of our function \(f(x)={(x-h)}^2+c\) is the same. Let’s say that we start at the point \(x=h\). If we now increase the value of \(x\) with \(a\) so that \(x=(h+a)\) we end up with:

$$f(h+a)={((h+a)-h)}^2+c$$

$$f(h+a)=a^2+c$$

If we instead decrease the value of \(x\) with \(a\) so that \(x=(h-a)\) we end up with:

$$f(h-a)={((h-a)-h)}^2+c$$

$$f(h-a)=a^2+c$$

Notice that both cases give us the same result. If we move \(a\) units to the right of the point where \(x=h\), we get the same value as if we had moved \(a\) units to the left of the point where \(x=h\). The graph must therefore be symmetrical around \(x=h\).

The fact that quadratic equations are symmetrical can be used for more than finding the vertex, i.e. the maximum/minimum points. It can also be used to find the second solution if one solution and the symmetry line are known. This is because both solutions have the same \(y\) value (\(0\)) and therefore they must be the same \(x\) value away from the symmetry line. One solution is always to the left of the symmetry line and the other is to the right of the symmetry line.

This symmetry can be seen in the abc formula.

The abc formula is \( x = - \frac {b}{2a} \pm \frac {\sqrt {b^2-4ac}} {2a} \)

\( x = - \frac {b}{2a} \) is the symmetry line. \( \frac {\sqrt {b^2-4ac}} {2a} \) is the distance left and right from the symmetry line where the solutions are. This is because the graph intersects the \(x\)-axis in the two points \( x = - \frac {b}{2a} + \frac {\sqrt {b^2-4ac}} {2a} \) and \( x = - \frac {b}{2a} - \frac {\sqrt {b^2-4ac}} {2a} \) which are equally far from \( x = - \frac {b}{2a}\).

If you graph a quadratic function, you will get one out of two types:

The first types are the quadratics with a positive coefficient \(a\) in front of the \(x^2\) term. These graphs go from \( +\infty \) downwards to a min point and then upwards to \( +\infty \). Some examples:

$$ y=x^2 $$

$$ y=2,5x^2-5x+16 $$

The second type are the quadratics with a negative coefficient \(a\) in front of the \(x^2\) term.These graphs go from \( -\infty \) upwards to a max point and then downwards to \( -\infty\).

Some examples:

$$ y=-x^2 $$

$$ y=-3x^2-7x+9 $$

An easy rule for remembering these two types is that the quadratic with the positive \(x^2\) coefficient gives a happy face while the quadratic with the negative \(x^2\) coefficient gives a sad face.

This fact raises a question. Why do these two types occur?

Let us find out.

To answer this question we need to break down the function into its different parts. This can be done in several forms but I will focus here on the abc form since it is the most commonly used.

$$ y=ax^2+bx+c $$

We can start by \(c\), since it is the easiest. \(c\) is always \(c\). It is a constant. If you change the value of \(c\), all that happens is that the entire graph shifts upwards or downwards. Therefore, \(c\) has nothing to do with the form of the graph.

What role \(b\) has is actually quite hard to find out. It is one of two constants that determine the vertex (together with \(a\)) and it changes how wide the graph is. With a greater b, the symmetry line moves to the left. With a greater \( \lvert b \rvert \) the distance between the symmetry line and the roots increases. What we need to know here is that no matter which \(b\) we pick, the graph will have the same overall shape (happy face or sad face).

So neither \(c\) nor \(b\) can turn the graph upside down. The perpetrator must therefore be \(a\). More specifically, if \(a>0\) (positive) then the graph becomes a happy face and if \(a0\), then \(ax^2\) will be a very large number, closer and closer to \( + \infty \), and we will therefore get a happy-faced graph. The opposite goes for \( a0 \Leftrightarrow \text { min point} $$

$$ a

Vieta’s formulas, or Viète’s laws, are used to relate the solutions to an equation with its coefficients. They can be used even when the equation is not quadratic but I will only describe how to use them in quadratics in this lesson.

An equation written on the form \(x^2+bx+c=0\) with the solutions \(x=p\) and \(x=q\) can also be expressed as \( (x-p)(x-q)=0 \). If we expand this expression we end up with:

$$ x^2+bx+c = x^2-(p+q)x+pq $$

This implies that:

$$ b = -(p+q) $$

$$ c = pq $$

These two equations are Vieta’s formulas.

If \( a \ne 1\) then the same applies but with \( \frac {b}{a} \) and \( \frac {c}{a} \) instead of \(b\) and \(c\) because we divide by \(a\) to get it in the right form.

Vieta’s formulas have many applications, and here I will state some of them. You will surely find more when working with more quadratic equations. Some of the things we can use Vieta’s formulas for include:

1) Finding the second solution when we know the first one. We do this by replacing \(b\) and \(c\) (or \( \frac {b}{a} \) and \( \frac {c}{a} \)) with the right constants and replacing \(p\) with our known solution. From there we can solve the equation and get the second solution \(q\).

2) Control if your answer is right. Use the two solutions you have got in the formulas and see if you get the equation you started with.

3) Find the solutions faster without having to use the abc formula. Note: This is not always possible, and in exams, you will have to write more than just the solutions.

There are certain equations which do not look like quadratic equations from the start but which can be transformed into quadratics. Once transformed, we can solve them with methods that we have learned. However, because the original equation was not quadratic there is a risk that we end up with false solutions. Such false solutions are called extraneous solutions. Let’s look at an example of such an equation.

$$ \sqrt{2-x} = x $$

The only solution to the root equation is \(x=1\) even though the quadratic equation had two solutions. This is because when a negative number is squared, the result is positive. Therefore, the “negativeness” gets lost when solving root equations this way and hence the extraneous root. It is not always easy to know if there are any extraneous roots and, in that case, which one it is. Therefore, always test both solutions in the original equation when solving root equations.

Apart from equations involving square roots, there is also one more form of equation which can be transformed to a quadratic equation: equations with denominators. And similar to the root equation, you have to double check your solutions because it may be a solution which works with the quadratic equation but not with the original one.

To solve equations with denominators, you have to multiply both sides with the product of the denominators. Then you solve the quadratic equation that forms. After that you check if there are any extraneous roots. But let us take it step by step with an example:

Find all the solutions to the equation \( \frac{x}{x-1} - \frac{2}{x} = \frac{1}{x-1} \)

Two important concepts to keep in mind when dealing with quadratic equations are implication and equivalence, and the differences between the two. These are logic operators which are used in mathematics in different areas. One of the most common is when proving theorems. However, it is good to know the difference when dealing with square roots.

Implication means that \(A\) leads to \(B\). If you know \(A\) then you know \(B\) as well. This is symbolised with an arrow, as \( A \Rightarrow B\). The implication can go in the other way too. You can write that \(A\) leads to \(B\) both \( A \Rightarrow B\) and \( B \Leftarrow A\). In both cases the arrow points from \(A\) to \(B\). Here is an example of implication:

$$ \text{ It is raining. } \Rightarrow \text{ It is cloudy.} $$

If it is raining, it always has to be cloudy. The arrow does not point in the other way too because if we only know that it is cloudy, it does not have to be raining.

Equivalence is when the arrow points both ways. \(C\) is equivalent with \(D\) is written \( C \Leftrightarrow D \) and it is the same thing as \( D \Leftrightarrow C \). An example of equivalence is:

$$ \text{ It is July. } \Leftrightarrow \text{ Next month is August.} $$

If we only know that it is July, the next month has to be August. And if we only know that the next month is August we are equally sure that it is July right now.

One can use these arrows with mathematical statements as well and when dealing with quadratic equations one important thing to have in mind is which arrow to use here:

Look at the two statements \(A\) and \(B\) and replace the question mark with the right arrow (\( \Rightarrow \), \( \Leftarrow \) or \( \Leftrightarrow \))

$$ A: x=3 $$

$$ B: x^2=9 $$

$$ A \: ? \: B $$

Simplifying radicals like \( \sqrt{5+ \sqrt{24}} \) can be useful when working with quadratic equations (and for many other problems too). The most obvious example is when solving equations like \( x^2 = 5+ \sqrt{24} \). \( \sqrt{5+ \sqrt{24}} \) is not a very simplified expression and therefore it is good to know how to simplify it. (And it is very satisfying to do it.)

We do this by guessing that the expression can be written as \( \sqrt{x} + \sqrt{y} \). This implies that

$$ \sqrt{x} + \sqrt{y} =  \sqrt{5+ \sqrt{24}} $$

$$ x + 2 \sqrt{xy} + y = 5 + \sqrt{24} $$

so

$$ x + y = 5 $$

$$ 2 \sqrt{xy} = \sqrt{24} $$

$$ 4xy = 24 $$

$$ xy = 6 $$

Now we have a system of equations: \(x+y=5\) and \(x \cdot y=6\).

In this case, we can just try it and then find that \(2\) and \(3\) fits. Otherwise one can solve it for example by substituting \(y\) for \((5-x)\) in the equation \(x \cdot y=6\) and solving the equation that occurs.

No matter how we do it, in the end we get that \( \sqrt{5+ \sqrt{24}} = \sqrt{2} + \sqrt{3} \) and this looks a lot nicer.

Note: If we had had an equation such as \( \sqrt{5 - \sqrt{24}} \), we would have had to guess that it could be written as \( \sqrt{x} - \sqrt{y} \) instead.

We can use the same method to solve equations on the form \( \sqrt{5-12i} \) by guessing that the expression can be written as \(a+bi\). This implies that:

$$ {a+bi}^2 = {\sqrt{5-12i}}^2 $$

$$ a^2 + 2abi + {(bi)}^2 = 5 - 12i $$

so

$$ a^2 - b^2 = 5 $$

$$ 2ab = 12 $$

or

$$ a^2 - b^2 = 5 $$

$$ ab = 6 $$

Which we can solve, for example by substituting \(b\) for \( \frac{6}{a} \) in the equation \(a^2-b^2=5\) or by writing it as \( (a+b)(a-b)=5 \) and \(ab=6\).

In the same way as before, no matter how we do it, we will find that \( \sqrt{5-12i} =3+2i\).