Quadratic Equations

Hello students, welcome to today's mathematics lesson. I'm so happy to see you all here, ready to learn about one of the most important topics in algebra - quadratic equations. This chapter is going to be extremely useful for you, not just for your exams but also for understanding many real-life situations. So let's begin our journey into the world of quadratic equations.

In Chapter 2, you studied different types of polynomials. You remember that a polynomial of degree 2 is called a quadratic polynomial, and it has the form ax² + bx + c, where a is not equal to zero. Well, today we are going to learn what happens when we equate this polynomial to zero. When we set ax² + bx + c equal to zero, we get a quadratic equation. And these equations appear everywhere in real life - in architecture, in business, in physics, and in many other fields.

Let me give you a real-life example from the chapter itself. Suppose a charity trust decides to build a prayer hall with a carpet area of 300 square metres, and the length is one metre more than twice its breadth. How would we find the length and breadth? Well, let's say the breadth is x metres. Then the length would be 2x + 1 metres. The area would be length times breadth, which is (2x + 1) multiplied by x, giving us 2x² + x square metres. Since we know the area is 300 square metres, we get the equation 2x² + x = 300. Rearranging this, we get 2x² + x - 300 = 0. This, students, is a quadratic equation! So you see, quadratic equations help us solve practical problems like this one.

Now let me tell you a bit about the history of quadratic equations. Many people believe that the Babylonians were the first to solve quadratic equations. They knew how to find two positive numbers when we know their sum and their product, which is exactly equivalent to solving a quadratic equation of the form x² - px + q = 0. The Greek mathematician Euclid developed a geometrical approach for finding lengths that satisfy quadratic equations. But solving quadratic equations in the general form is often credited to ancient Indian mathematicians. The great mathematician Brahmagupta, who lived around 598 to 665 CE, gave an explicit formula to solve a quadratic equation of the form ax² + bx = c. Later, Sridharacharya, who lived around 1025 CE, derived the famous quadratic formula that we still use today - this was quoted by Bhaskara II. An Arab mathematician named Al-Khwarizmi, who lived around 800 CE, also studied different types of quadratic equations. And in Europe, Abraham bar Hiyya Ha-Nasi published a book in 1145 CE that gave complete solutions of various quadratic equations. So you see, quadratic equations have a rich and fascinating history across many civilizations!

Now let's understand exactly what a quadratic equation is. A quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, and c are real numbers, and importantly, a is not equal to zero. We say a cannot be zero because if a were zero, the equation would no longer be quadratic - it would become linear. For example, 2x² + x - 300 = 0 is a quadratic equation. Similarly, 2x² - 3x + 1 = 0, 4x - 3x² + 2 = 0, and 1 - x² + 300 = 0 are also quadratic equations. Notice that in the third example, the terms are not in descending order of degree. When we write the terms in descending order of their degrees, we get -3x² + 4x + 2 = 0, which is of the form ax² + bx + c = 0 with a = -3, b = 4, and c = 2. This standard form - ax² + bx + c = 0 with a not equal to zero - is what we call the standard form of a quadratic equation.

Let me now show you how to represent real-life situations as quadratic equations. This is Example 1 from your textbook, and it's very important that you understand this process thoroughly.

Consider the first situation. John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We need to find how many marbles they had to start with.

Let's solve this step by step. Let the number of marbles John had be x. Then, since together they have 45 marbles, Jivanti must have 45 - x marbles. Now, both of them lost 5 marbles each. So John now has x - 5 marbles, and Jivanti now has 45 - x - 5, which is 40 - x marbles. The product of the marbles they now have is given as 124. So we have (x - 5) multiplied by (40 - x) equals 124. Let's expand this: (x - 5)(40 - x) = 40x - x² - 200 + 5x = -x² + 45x - 200. So we get -x² + 45x - 200 = 124. Bringing 124 to the left side, we get -x² + 45x - 324 = 0. Multiplying both sides by -1, we get x² - 45x + 324 = 0. And there we have it - a quadratic equation! This represents the problem mathematically.

Now let's look at the second situation. A cottage industry produces a certain number of toys in a day. The cost of production of each toy was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹750. We need to find the number of toys produced that day.

Let the number of toys produced be x. Then, the cost of production of each toy is 55 - x rupees. So the total cost would be x multiplied by (55 - x), which is 55x - x². This equals 750. So we have 55x - x² = 750. Rearranging, we get -x² + 55x - 750 = 0, or multiplying by -1, we get x² - 55x + 750 = 0. Again, we have a quadratic equation!

So students, you can see that many real-life situations can be represented as quadratic equations. The key is to identify the unknown quantity, express other quantities in terms of it, and then form the equation.

Now let's look at Example 2, where we check whether given equations are quadratic equations. This is very important because sometimes an equation may look like a quadratic equation at first glance, but after simplification, it might turn out to be something else. Or sometimes, an equation that doesn't look quadratic might actually be quadratic after simplification!

Let's check the first one: (x - 2)² + 1 = 2x - 3. We need to expand the left-hand side and simplify. (x - 2)² is x² - 4x + 4, so adding 1 gives us x² - 4x + 5. So the equation becomes x² - 4x + 5 = 2x - 3. Bringing all terms to one side, we get x² - 4x + 5 - 2x + 3 = 0, which is x² - 6x + 8 = 0. This is of the form ax² + bx + c = 0, so yes, it is a quadratic equation.

Now the second one: x(x + 1) + 8 = (x + 2)(x - 2). Let's simplify both sides. The left side is x² + x + 8. The right side is (x + 2)(x - 2), which is x² - 4. So we have x² + x + 8 = x² - 4. Subtracting x² from both sides, we get x + 8 = -4, or x + 12 = 0. This is not of the form ax² + bx + c = 0 because there is no x² term. So this is NOT a quadratic equation. Students, notice how this equation appears to involve x² on both sides, but after simplification, the x² terms cancel out! This is why we must always simplify before deciding.

The third one: x(2x + 3) = x² + 1. The left side is 2x² + 3x. So we have 2x² + 3x = x² + 1. Subtracting x² from both sides, we get x² + 3x - 1 = 0. This is of the form ax² + bx + c = 0, so it IS a quadratic equation.

The fourth one: (x + 2)³ = x³ - 4. Now, (x + 2)³ expands to x³ + 6x² + 12x + 8. So we have x³ + 6x² + 12x + 8 = x³ - 4. Subtracting x³ from both sides, we get 6x² + 12x + 8 = -4, or 6x² + 12x + 12 = 0. Dividing by 6, we get x² + 2x + 2 = 0. This is of the form ax² + bx + c = 0, so it IS a quadratic equation! Students, notice how this equation looks like a cubic equation (degree 3) because of the cubes, but after simplification, it reduces to a quadratic equation. This is a very important observation.

So here's the key takeaway: always simplify the given equation first before deciding whether it is quadratic or not. Sometimes the equation might appear to be quadratic but isn't, and sometimes it might appear to be something else but actually is quadratic.

Now let's move on to Section 4.3, where we learn how to find the roots of a quadratic equation by factorisation. But first, let me explain what we mean by a root or solution of a quadratic equation.

Consider the quadratic equation 2x² - 3x + 1 = 0. If we put x = 1 in the left-hand side, we get (2 × 1²) - (3 × 1) + 1 = 2 - 3 + 1 = 0, which equals the right-hand side. So we say that 1 is a root of this quadratic equation. In general, a real number α is called a root of the quadratic equation ax² + bx + c = 0 if aα² + bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the equation. An important point to remember is that the zeroes of the quadratic polynomial ax² + bx + c and the roots of the quadratic equation ax² + bx + c = 0 are exactly the same. Since a quadratic polynomial can have at most two zeroes (as you learned in Chapter 2), any quadratic equation can have at most two roots.

Now let's learn how to find these roots by factorisation. You already learned in Class IX how to factorise quadratic polynomials by splitting the middle term. We will use this same technique to find the roots.

Let's look at Example 3: Find the roots of 2x² - 5x + 3 = 0 by factorisation.

We need to split the middle term -5x into two terms such that their product equals the product of the coefficient of x² (which is 2) and the constant term (which is 3). So we need two numbers whose product is 2 × 3 = 6 and whose sum is -5. Those numbers are -2 and -3, because -2 × -3 = 6 and -2 + -3 = -5. So we write -5x as -2x - 3x.

So 2x² - 5x + 3 = 2x² - 2x - 3x + 3. Now we factorise by grouping: take 2x common from the first two terms: 2x(x - 1). Take -3 common from the last two terms: -3(x - 1). So we have 2x(x - 1) - 3(x - 1) = (2x - 3)(x - 1).

So the equation 2x² - 5x + 3 = 0 can be written as (2x - 3)(x - 1) = 0. Now, if the product of two factors is zero, then either the first factor is zero or the second factor is zero (or both). So we have 2x - 3 = 0 or x - 1 = 0. Solving these, we get x = 3/2 or x = 1. So the roots are 3/2 and 1. We can verify this by substituting back into the original equation. For x = 1, we get 2(1)² - 5(1) + 3 = 2 - 5 + 3 = 0. For x = 3/2, we get 2(3/2)² - 5(3/2) + 3 = 2(9/4) - 15/2 + 3 = 9/2 - 15/2 + 3 = -6/2 + 3 = -3 + 3 = 0. Both satisfy the equation. So students, this is the process of finding roots by factorisation.

Let's do another example. Example 4: Find the roots of 6x² - x - 2 = 0.

We need to split -x into two terms such that their product is 6 × (-2) = -12, and their sum is -1. The numbers 3 and -4 work, because 3 × (-4) = -12 and 3 + (-4) = -1. So we write -x as +3x - 4x.

So 6x² - x - 2 = 6x² + 3x - 4x - 2. Now factorise: take 3x common from the first two terms: 3x(2x + 1). Take -2 common from the last two terms: -2(2x + 1). So we have 3x(2x + 1) - 2(2x + 1) = (3x - 2)(2x + 1).

So the equation becomes (3x - 2)(2x + 1) = 0. Therefore, 3x - 2 = 0 or 2x + 1 = 0. Solving these, we get x = 2/3 or x = -1/2. So the roots are 2/3 and -1/2. We can verify these by substituting back into the original equation.

Now let's look at a slightly more complicated example. Example 5: Find the roots of 3x² - 2√6x + 2 = 0.

This looks a bit scary because of the square root symbol, but don't worry - we can handle it the same way. We need to split -2√6x into two terms such that their product is 3 × 2 = 6, and we need to use √6 in some way. Notice that (√6)² = 6. So let's try splitting the middle term as -√6x - √6x. Then 3x² - √6x - √6x + 2. Now factorise: from the first two terms, take √3x common: √3x(√3x - √2). From the last two terms, take -√2 common: -√2(√3x - √2). So we have (√3x - √2)(√3x - √2), which is (√3x - √2)².

So the equation becomes (√3x - √2)² = 0, which means √3x - √2 = 0. Solving this, we get √3x = √2, so x = √2/√3 = √(2/3). Now here's the interesting part - this root is repeated twice! We get the same root from both factors. So the equation has two equal roots, both equal to √(2/3). This is what we call a repeated root or a double root.

So students, when we factorise a quadratic equation and get the same factor twice, we get a repeated root. This is an important observation that will connect to what we learn about the nature of roots.

Now let's move on to Section 4.4, where we learn about the nature of roots and the quadratic formula. This is a very important section because it tells us how many roots a quadratic equation has and what kind of roots they are - without actually solving the equation!

You already know the quadratic formula for solving ax² + bx + c = 0. The roots are given by x = (-b ± √(b² - 4ac)) / 2a. This formula works for any quadratic equation, as long as the expression under the square root is non-negative. Now, the expression under the square root, b² - 4ac, is called the discriminant of the quadratic equation. The word "discriminant" means something that helps us distinguish or decide. And indeed, the discriminant helps us determine the nature of the roots without solving the equation.

Let's see how. If b² - 4ac > 0, that is, if the discriminant is positive, then we get two distinct real roots: (-b + √(b² - 4ac)) / 2a and (-b - √(b² - 4ac)) / 2a. These are different from each other.

If b² - 4ac = 0, that is, if the discriminant is zero, then both roots become (-b / 2a). So we get two equal real roots - both are the same. We say the equation has two coincident or equal real roots in this case.

If b² - 4ac < 0, that is, if the discriminant is negative, then there is no real number whose square is b² - 4ac (since the square of any real number is non-negative). Therefore, there are no real roots in this case. The roots are complex or imaginary numbers, which we will study in higher classes.

So the discriminant b² - 4ac is the key to understanding the nature of roots. A quadratic equation ax² + bx + c = 0 has two distinct real roots if b² - 4ac > 0, two equal real roots if b² - 4ac = 0, and no real roots if b² - 4ac < 0.

Now let's look at some examples to understand this better.

Example 7: Find the discriminant of the quadratic equation 2x² - 4x + 3 = 0, and hence find the nature of its roots.

The given equation is in the form ax² + bx + c = 0, where a = 2, b = -4, and c = 3. Now, the discriminant b² - 4ac = (-4)² - 4 × 2 × 3 = 16 - 24 = -8, which is less than 0. So the discriminant is negative. Therefore, the equation has no real roots. That's it - we didn't even need to solve the equation to know this! The discriminant tells us that the roots are not real.

Now let's look at a more interesting example. Example 8: A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

This is a word problem that involves geometry and quadratic equations. Let's understand the problem. We have a circular park with diameter 13 metres. There are two gates A and B that are diametrically opposite each other - that means they are on opposite ends of a straight line passing through the center of the circle. So the distance AB is 13 metres. We need to place a pole at a point P on the boundary of the park such that the difference between its distances from A and B is 7 metres.

Let the distance of the pole from gate B be x metres. Then the distance from the pole to gate A would be x + 7 metres (since the difference is 7 metres). Now, since A and B are diametrically opposite, the angle APB is a right angle (this is a property of circles - the angle subtended by a diameter is a right angle). So we have a right triangle APB with right angle at P. By the Pythagoras theorem, AP² + PB² = AB². So we have (x + 7)² + x² = 13².

Let's expand this: x² + 14x + 49 + x² = 169. So 2x² + 14x + 49 = 169. Simplifying, we get 2x² + 14x - 120 = 0. Dividing by 2, we get x² + 7x - 60 = 0.

Now, is it possible to place the pole? This depends on whether this quadratic equation has real roots. Let's check the discriminant. For x² + 7x - 60 = 0, we have a = 1, b = 7, c = -60. So the discriminant b² - 4ac = 7² - 4 × 1 × (-60) = 49 + 240 = 289, which is greater than 0. So the equation has two real roots, which means it IS possible to place the pole on the boundary of the park.

Now let's find the actual distances. Using the quadratic formula, x = (-7 ± √289) / 2 = (-7 ± 17) / 2. So we have two possibilities: x = (-7 + 17) / 2 = 10/2 = 5, or x = (-7 - 17) / 2 = -24/2 = -12. Since x represents a distance, it must be positive. So we ignore x = -12. Therefore, x = 5 metres. This means the pole should be placed 5 metres from gate B. Then the distance from gate A would be x + 7 = 5 + 7 = 12 metres. So the pole should be erected at a distance of 5 metres from one gate and 12 metres from the other gate.

This is a wonderful example of how quadratic equations can be used to solve real-world geometry problems!

Now let's look at Example 9: Find the discriminant of the equation 3x² - 2x + 1/3 = 0 and hence find the nature of its roots. Find them if they are real.

Here a = 3, b = -2, and c = 1/3. The discriminant b² - 4ac = (-2)² - 4 × 3 × (1/3) = 4 - 4 = 0. So the discriminant is zero. Therefore, the equation has two equal real roots. The roots are given by -b/2a = -(-2) / (2 × 3) = 2/6 = 1/3. So both roots are equal to 1/3. We can verify this: substituting x = 1/3 into the equation, we get 3(1/3)² - 2(1/3) + 1/3 = 3(1/9) - 2/3 + 1/3 = 1/3 - 2/3 + 1/3 = 0. Yes, it works!

So students, let's recap what we've learned in this chapter.

First, we learned that a quadratic equation in the variable x is an equation of the form ax² + bx + c = 0, where a, b, c are real numbers and a is not equal to zero. This is the standard form of a quadratic equation.

Second, we learned that a real number α is called a root of the quadratic equation ax² + bx + c = 0 if aα² + bα + c = 0. The zeroes of the quadratic polynomial and the roots of the quadratic equation are the same.

Third, we learned that if we can factorise ax² + bx + c into a product of two linear factors, then we can find the roots by equating each factor to zero. This is called solving by factorisation.

Fourth, we learned the quadratic formula, which gives the roots of any quadratic equation: x = (-b ± √(b² - 4ac)) / 2a, provided b² - 4ac is greater than or equal to zero.

Fifth, and most importantly, we learned about the discriminant. The discriminant is b² - 4ac, and it tells us the nature of the roots without solving the equation. If b² - 4ac > 0, the equation has two distinct real roots. If b² - 4ac = 0, the equation has two equal real roots. If b² - 4ac < 0, the equation has no real roots.

We also saw how quadratic equations arise in many real-life situations - in finding dimensions of rectangular plots, in problems involving ages, in geometry problems involving circles, and so on. The key is to translate the word problem into a mathematical equation, and then solve it using either factorisation or the quadratic formula.

Students, quadratic equations are fundamental to mathematics and appear in many topics you will study in the future, including physics, economics, and statistics. So make sure you understand this chapter thoroughly. Practice as many problems as you can, and don't hesitate to go back and review any concept that you find confusing.

That's all for today's lesson. Thank you for your attention and patience. Keep practicing, and remember - mathematics is not about memorisation, it's about understanding and applying concepts. Good luck with your studies!