Quadratic Factorisation & Equation

Quadratic expressions are polynomials of order $2$. Any expression of the form $ax^2 + bx + c$, where $a,\ b\ and\ c$ are constants is called a quadratic expression.

Today we will learn a simple method to factorise a quadratic expression and subsequently how to solve a quadratic equation of the form:

$ax^2 + bx + c = 0$

I. Quadratic Factorisation:

In this method we try to split the middle term, that is the $x$ term such that the product of the two splits is equal to the product of the first and last terms. In other word we write $bx$ as $b_1x + b_2x$ such that, $b_1x + b_2x = bx$ and $b_1x \times b_2x = ax^2 \times c$. We will take some examples to understand this:

Let us factorise the expression $3x^2 + 4x - 15$

Here, $a = 3$, $b = 4$ and $c = -15$

The product of the first and the last term is $3x^2 \times (-15) = -45x^2$

Let us try to split up the term $+4x$ into two parts such that the product is $-45x^2$

We will first check how many different ways can $45$ be factorised into two factors, and we will need to see the sum and the difference of the two factors.

$45 = 45 \times 1$, sum of factors = $46$, difference = $44$

$45 = 3 \times 15$, sum of factors = $18$, difference = $12$

$45 = 9 \times 5$ , sum of factors = $14$, difference = $4$

Now we see that $4$ is same as the coefficient of the $x$ term.

So, if we split up $4x$ as $9x - 5x$, we get $(9x) \times (-5x) = -45x^2$ and this is what we are looking for.

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So, we can write our expression as:

$3x^2 + 4x - 15$

$= 3x^2 + 9x - 5x - 15$ (Now take $3x$ common from the first pair of terms and $-5$ from the second pair)

$= 3x(x + 3) -5(x+3)$ (Now we can take $x+3$ common from the whole expression)

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

Now we have our expression as a product of two factors.

Let us take one more example to understand this better. We will try to factorise the expression $4x^2 - 8x - 5$

Like before, we will check the possible factors of $20$ and pick the ones whose sum or difference is $-8$

$20 = 20 \times 1$, sum = $21$, difference = $19$

$20 = 4 \times 5$, sum = $9$, difference = $1$

$20 = 10 \times 2$, sum = $12$, difference = $8$

hence we pick $10 \times 2$ and factorise our expression as shown below:

$4x^2 - 8x - 5$

$= 4x^2 - 10x + 2x - 5$

$= 2x(2x - 5) + 1(2x - 5)$

$= (2x - 5)(2x + 1)$

Let us take one more example:

$= x^2 - 11x + 30$, we can see that $30 = 5 \times 6$ and $(-5) + (-6) = -11$

$= x^2 - 6x - 5x + 30$

$= x(x - 6) - 5(x - 6)$

$= (x - 6)(x - 5)$

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You might have observed by now, that if the product of the $x^2$ and the constant term is a positive number, our solution will require the sum of the two middle terms and if the product is a negative number then our solution will require the difference of the two middle terms.

II. Solving Quadratic Equations

Now that we know how to factorise a quadratic expression, it should be pretty straight forward for us to understand how to solve a quadratic equation of the form $ax^2 + bx + c = 0$

Let us say, we can factorise the quadratic expression $ax^2 + bx + c$ as $(px + q)(rx + s)$ then we can write our equation as:

$ax^2 + bx + c = 0$

$\texttip{\Rightarrow}{follows that} (px + q)(rx + s) = 0$

Now, we know that if the product of two numbers is $0$ then either one of them is $0$ or both are zero.

Therefore, in our expression either $(px + q) = 0$ or $(rx + s) = 0$

If $(px + q) = 0$ then $x = -\dfrac{p}{q}$

or, if $(rx + s) = 0$ then $x = -\dfrac{s}{r}$

Quadratic equations will normally have two solutions, unless both the factors are exactly same, in that case our quadratic expression will be a perfect square.

Now we will solve all our previous expression as equation and see what solutions we get.

$\underline{Example\ 1:}$

$3x^2 + 4x - 15 = 0$

$\texttip{\Rightarrow}{follows that} (x+3)(3x-5) = 0$

$\texttip{\Rightarrow}{follows that} x + 3 = 0$ or $3x - 5 = 0$

$\texttip{\therefore}{therefore} x = -3$ or $x = \dfrac{5}{3}$

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$\underline{Example\ 2:}$

$4x^2 - 8x - 5 = 0$

$\texttip{\Rightarrow}{follows that} (2x - 5)(2x + 1) = 0$

$\texttip{\Rightarrow}{follows that} 2x - 5 = 0$ or $2x + 1 = 0$

$\texttip{\therefore}{therefore} x = \dfrac{5}{2}$ or $x = -\dfrac{1}{2}$

Now let us try to solve a problem using this concept:

$\underline{Example\ 3:}$

You have $4$ books more than your friend. The product of the number of books that the two of you have is $165$. How many books do you have?

Let us say you have $x$ books, therefore, your friend has $x - 4$

As per the given condition $x(x-4) = 165$.

Let us solve this equation using the steps below:

$x(x-4) = 165$

$\texttip{\Rightarrow}{follows that} x^2 - 4x = 165$

$\texttip{\Rightarrow}{follows that} x^2 - 4x - 165 = 0$

$\texttip{\Rightarrow}{follows that} x^2 - 15x + 11x - 165 = 0$

$\texttip{\Rightarrow}{follows that} x(x - 15) + 11(x - 15) = 0$

$\texttip{\Rightarrow}{follows that} (x - 15)(x + 11) = 0$

$\texttip{\therefore}{therefore} x = 15$ or $x = -11$

The number of books that you have has to be a positive number. Therefore, we can eliminate the solution $x = -11$ and the correct answer will be $15$

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III. Po-Shen Loh Method Of Solving Quadratic Equation:

This a relatively new method of solving quadratic equation. This method was discovered by Po-Shen Loh

Given a quadratic equation of the form $ax^2 + bx + c = 0$, for many cases we can use this method to find the roots very fast.

This method relies on the fact that the sum of the two roots of a quadratic equation is $-\dfrac{b}{a}$ and the product of the two roots is $\dfrac{c}{a}$

Now, if we take the two roots as $-\dfrac{b}{2a} - u$ and $-\dfrac{b}{2a} + u$ then the sum of the roots is automatically becoming $-\dfrac{b}{a}$.

So, knowing $a$, $b$ and $c$, we need to solve the equation:
$\left( -\dfrac{b}{2a} - u \right)\left( -\dfrac{b}{2a} - u \right) = \dfrac{c}{a}$

The concept may appear a little complex but it is actually a very simple method which we will see with the following example.

Let's consider the equation:

$2x^2 + 39x + 180 = 0$

The sum of the roots is $-\dfrac{39}{2}$ and the product of the roots is $\dfrac{180}{2} = 90$

Let us take the two roots as $\left( -\dfrac{39}{4} - u\right)$ and $\left( -\dfrac{39}{4} + u\right)$

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Considering the product of the two roots, we get:
$\left( -\dfrac{39}{4} - u\right)\left( -\dfrac{39}{4} + u\right) = 90$

$\Rightarrow \dfrac{39^2}{4^2} - u^2 = 90$

$\Rightarrow u^2 = \dfrac{39^2}{4^2} - 90$

$\Rightarrow u^2 = \dfrac{39^2 - 90\times 4^2}{4^2} = \dfrac{3^2(13^2 - 10 \times 4^2}{4^2} = \dfrac{3^2(169 - 160)}{4^2}$

$\Rightarrow u^2 = \dfrac{3^2(9)}{4^2} = \dfrac{3^2 \times 3^2}{4^2}$

$\Rightarrow u = \dfrac{9}{4}$ (Notice that we considered only the positive root, but we will get the same result for the negative root as well)

Therefore, the two roots are:

$\left( -\dfrac{39}{4} - \dfrac{9}{4}\right)$ and $\left( -\dfrac{39}{4} + \dfrac{9}{4}\right)$

that is:

$=-\dfrac{48}{4}$ and $-\dfrac{30}{4}$

$= -12$ and $-\dfrac{15}{2}$

This method can be applied for quadratic equations which have slightly larger values of $a$, $b$ and $c$.