Welcome to today's mathematics lesson. We are going to explore Chapter Five: Quadratic Equations. By the end of this session, you will understand what makes an equation quadratic, how to identify one, how to solve quadratic equations using factorisation and the quadratic formula, how to examine the nature of roots without actually solving, and how to tackle equations that can be reduced to quadratic form.
Let us begin with the fundamental question: what exactly is a quadratic equation? A quadratic equation is an equation in one variable where the highest power of that variable is two. In other words, the variable is squared, but we do not see any higher powers like cubes or fourth powers.
The standard form of any quadratic equation is ax² + bx + c = 0, where a, b, and c are real numbers, and crucially, a must not equal zero. If a were zero, the x² term would disappear, and we would no longer have a quadratic equation.
Quadratic equations come in two flavours. When an equation contains only the square term and a constant, like x² = 4 or 3x² – 8 = 0, we call it a pure quadratic equation. But when it contains both the square term and the first power term, like 4x² + 5x = 0, we call it an adfected quadratic equation.
How do we verify if a given equation is quadratic? We simplify it completely, clearing all brackets and fractions, and check whether it can be written in the standard form. For instance, if we expand (3x – 2)(2x – 3) = (2x + 5)(2x – 1), we get 6x² – 13x + 6 = 4x² + 8x – 5. Rearranging gives 2x² – 21x + 11 = 0, which is clearly quadratic.
Now let us turn to solving quadratic equations, and our first powerful method is factorisation. To solve a quadratic equation means to find the value or values of the variable that satisfy the equation. These values are called the roots or solutions of the equation.
The method of factorisation rests on a simple but profound principle called the Zero Product Rule. This rule states that whenever the product of two expressions equals zero, at least one of those expressions must be zero. So if we can write our quadratic as (x + 3)(x – 2) = 0, then either x + 3 = 0 or x – 2 = 0, giving us x = –3 or x = 2.
Let me walk you through the steps. First, clear any fractions and brackets. Second, transpose all terms to the left side to get the form ax² + bx + c = 0. Third, factorise the left-hand side expression. Finally, set each factor equal to zero and solve.
Consider the equation 2x² – 7x = 39. We rewrite this as 2x² – 7x – 39 = 0. To factorise, we split the middle term: 2x² – 13x + 6x – 39 = 0. Grouping gives x(2x – 13) + 3(2x – 13) = 0, which becomes (2x – 13)(x + 3) = 0. Therefore, x = 13/2 or x = –3.
Sometimes we encounter special cases. When there is no constant term, like x² = 5x, we get x² – 5x = 0, which factorises as x(x – 5) = 0, giving roots x = 0 and x = 5. When we have x² = 16, we can write x² – 16 = 0, which is a difference of squares: (x + 4)(x – 4) = 0, so x = ±4.
We can also work backwards. If we know the roots are –2 and 3, the equation must be (x + 2)(x – 3) = 0, which expands to x² – x – 6 = 0. There is an elegant shortcut: if roots are given, the equation is x² – (sum of roots)x + (product of roots) = 0. Here, sum is –2 + 3 = 1 and product is –2 × 3 = –6, confirming our equation.
Not all quadratics factorise nicely, so we need a universal method. This brings us to the quadratic formula, which works for every quadratic equation.
First, let me introduce the discriminant. For any quadratic equation ax² + bx + c = 0 where a is not zero, the expression b² – 4ac is called the discriminant, denoted by D. This single expression holds the key to understanding everything about the roots.
The quadratic formula states that the roots are given by x = (–b ± √(b² – 4ac))/2a, or equivalently x = (–b ± √D)/2a.
Let me show you where this formula comes from. Starting with ax² + bx + c = 0, we multiply by 4a to get 4a²x² + 4abx + 4ac = 0. We rewrite this as (2ax)² + 2(2ax)(b) + b² – b² + 4ac = 0. This becomes (2ax + b)² = b² – 4ac. Taking square roots: 2ax + b = ±√(b² – 4ac), and solving for x gives our formula.
Let us apply this to x² = 18x – 77. Rewriting: x² – 18x + 77 = 0. Here a = 1, b = –18, c = 77. The discriminant is (–18)² – 4(1)(77) = 324 – 308 = 16. So x = (18 ± 4)/2, giving x = 11 or x = 7.
For equations requiring decimal precision, like 3x² + 5x – 9 = 0, we find b² – 4ac = 25 + 108 = 133. Then x = (–5 ± √133)/6. Since √133 is approximately 11.533, we get x = 1.09 or x = –2.76 to two decimal places.
Now we come to one of the most powerful applications of the discriminant: examining the nature of roots without actually finding them. The value of D = b² – 4ac tells us everything.
When D = 0, the roots are real and equal. When D > 0, the roots are real and unequal. And when D < 0, the roots are imaginary, not real.
Remember, every rational and irrational number is a real number. But the square root of a negative number is imaginary. So √–4 or √–8 are imaginary, not real.
Consider 5x² – 6x + 7 = 0. The discriminant is 36 – 140 = –104, which is negative. Therefore, the roots are imaginary.
For x² + 6x + 9 = 0, the discriminant is 36 – 36 = 0, so the roots are real and equal. Indeed, this is (x + 3)² = 0, giving a repeated root x = –3.
For 2x² + 6x + 3 = 0, the discriminant is 36 – 24 = 12, which is positive but not a perfect square. The roots are real, unequal, and irrational.
We can use this to find unknown coefficients. If (4 + m)x² + (m + 1)x + 1 = 0 has equal roots, then (m + 1)² – 4(4 + m) = 0. This simplifies to m² – 2m – 15 = 0, giving m = 5 or m = –3.
Finally, let us explore equations that do not look quadratic at first glance but can be reduced to quadratic form. These include equations with higher even powers like x⁴, or equations involving roots and complex fractions.
Consider 2x⁴ – 5x² + 3 = 0. If we substitute y = x², this becomes 2y² – 5y + 3 = 0. Factorising: (2y – 3)(y – 1) = 0, so y = 1 or y = 3/2. Since y = x², we get x = ±1 or x = ±√6/2.
For equations with reciprocal structures, like √(x/(1–x)) + √((1–x)/x) = 2 1/6, where x ≠ 0 and x ≠ 1, we substitute y = √(x/(1–x)). Then the second term becomes 1/y, and we solve y + 1/y = 13/6, which leads to 6y² – 13y + 6 = 0.
Some problems require clever substitutions based on algebraic identities. If we know that (x + 1/x)² – (x – 1/x)² = 4, we can use this relationship to transform complex equations into manageable quadratic forms.
Let me now recap the key takeaways from this chapter.
First, a quadratic equation in one variable has the standard form ax² + bx + c = 0 with a ≠ 0. Second, we can solve quadratics by factorisation using the Zero Product Rule, or by the universal quadratic formula. Third, the discriminant D = b² – 4ac reveals the nature of roots: equal when D = 0, real and distinct when D > 0, and imaginary when D < 0. Fourth, equations with higher even powers or special structures can often be reduced to quadratic form through substitution. Fifth, knowing the roots allows us to reconstruct the equation using sum and product relationships. And finally, always verify your solutions by substituting back into the original equation.
Quadratic equations are fundamental to mathematics and appear throughout science, engineering, and economics. Mastering these techniques gives you powerful tools for problem-solving. Keep practising, stay curious, and remember that every complex problem can be broken down into manageable steps. Until next time, happy learning.