CBSE Class 10 Mathematics Quadratic Equations Notes Set 03

Introduction to Quadratic Equations

You must have come across different types of equations in mathematics such as linear equations, linear equation in two variables and many more. This chapter will introduce you to a new set of equations called quadratic equations. The polynomial of the form \( ax^2 + bx + c,\ a \neq 0 \) is a quadratic polynomial. When we equate this quadratic polynomial to zero, we get a quadratic equation i.e. \( ax^2 + bx + c = 0,\ a \neq 0 \).

An equation of the form \( ax^2 + bx + c = 0,\ a \neq 0 \) where \( a, b, c \in R \) is called a quadratic equation. The constants a, b, c are called the coefficients of equation. There will be two roots for a quadratic equation and can be found by solving the quadratic equation \( ax^2 + bx + c = 0 \). Roots are also called solutions or zeros of quadratic equations.

The study of quadratic equation is important because of its application to parabolas and other conic sections. In this chapter, we will study about the quadratic equations and various ways of finding their roots.

Standard Form of a Quadratic Equation

The standard form of quadratic equation is \( ax^2 + bx + c = 0,\ a \neq 0 \) where \( a, b, c \in R \). Quadratic equations, in general are of the following types:

1. When \( b = 0,\ c \neq 0 \) i.e., of the type \( ax^2 + c = 0 \), it becomes a pure quadratic equation.
2. When \( b \neq 0,\ c = 0 \) i.e., of the type \( ax^2 + bx = 0 \) or when \( b \neq 0,\ c \neq 0 \) i.e., of the type \( ax^2 + bx + c = 0 \), it is an adfected quadratic equation.

Mind it.
1. A quadratic equation has two and only two roots.
2. A quadratic equation cannot have more than two different roots.
3. If \( \alpha \) be a root of the quadratic equation \( ax^2 + bx + c = 0 \) then \( (x - \alpha) \) is a factor of quadratic equation \( ax^2 + bx + c = 0 \).

Sum and Product of the Roots of a Quadratic Equation

Let \( \alpha,\ \beta \) be the roots of a quadratic equation \( ax^2 + bx + c = 0,\ a \neq 0 \), then\[ \alpha + \beta = \frac{-b}{a} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} \]and\[ \alpha\beta = \frac{c}{a} = \frac{\text{constant term}}{\text{coefficient of } x^2} \]

Sign of the Roots

Mind it.
1. If the two roots \( \alpha \) and \( \beta \) be reciprocal to each other, then \( a = c \).
2. If the two roots \( \alpha \) and \( \beta \) be equal in magnitude and opposite in sign then \( b = 0 \).

Solved Examples

Example 1. Which of the following are quadratic equations, give reason: (i) \( 4x^2 - 5x + 13 = 0 \); (ii) \( x(x + 1) = x^2 + 5 \); (iii) \( 2x + \frac{5}{x} = x^2 \); (iv) \( x^2 + \frac{2}{x^2} = 2 \).
Answer:
(i) Since \( 4x^2 - 5x + 13 \) is of the form \( ax^2 + bx + c = 0 \), \( 4x^2 - 5x + 13 = 0 \) is a quadratic equation.
(ii) \( x(x + 1) = x^2 + 5 \Rightarrow x^2 + x = x^2 + 5 \), i.e., \( x - 5 = 0 \). It is not of the form \( ax^2 + bx + c = 0 \). Therefore, the given equation is not a quadratic equation.
(iii) \( 2x + \frac{5}{x} = x^2 \Rightarrow 2x^2 + 5 = x^3 \Rightarrow x^3 - 2x^2 - 5 = 0 \), which is cubic and not a quadratic equation.
(iv) \( x^2 + \frac{2}{x^2} = 2 \Rightarrow x^4 + 2 = 2x^2 \Rightarrow x^4 - 2x^2 + 2 = 0 \), which is biquadratic and not a quadratic equation.

Example 2. Determine whether the given value of x is the solution of the given equation or not: \( (3x + 8)(2x + 5) = 0;\ x = 2\frac{2}{3} \).
Answer:
Put \( x = 2\frac{2}{3} = \frac{8}{3} \).
\( (3x + 8)(2x + 5) = \left(3 \times \frac{8}{3} + 8\right)\left(2 \times \frac{8}{3} + 5\right) = (8 + 8)\left(\frac{16}{3} + 5\right) = 16 \times \frac{31}{3} = \frac{496}{3} \neq 0 \).
Therefore the given value of x is not the solution of the given equation.

Example 3. In each of the following, determine whether the given values are solutions (roots) of the equation or not: (i) \( 4x^2 - 2x - 2 = 0;\ x = 1 \); (ii) \( x^2 + \sqrt{2}x - 4 = 0;\ x = \sqrt{2},\ x = -2\sqrt{2} \).
Answer:
(i) For \( x = 1 \), L.H.S. \( = 4x^2 - 2x - 2 = 4(1)^2 - 2(1) - 2 = 4 - 2 - 2 = 0 = \) R.H.S. Therefore \( x = 1 \) is a solution of the given equation.
(ii) For \( x = \sqrt{2} \), L.H.S. \( = (\sqrt{2})^2 + \sqrt{2}(\sqrt{2}) - 4 = 2 + 2 - 4 = 0 = \) R.H.S. Therefore \( x = \sqrt{2} \) is a solution of the given equation.
For \( x = -2\sqrt{2} \), L.H.S. \( = (-2\sqrt{2})^2 + \sqrt{2}(-2\sqrt{2}) - 4 = 8 - 4 - 4 = 0 = \) R.H.S. Therefore \( x = -2\sqrt{2} \) is also a solution of the given equation.

Example 4. The sum of a fraction and 7 times its reciprocal is \( \frac{45}{96} \). Represent the above statement in the form of a quadratic equation.
Answer:
Let the fraction be x. Then, according to the question, \( x + \frac{7}{x} = \frac{45}{96} \Rightarrow 96x^2 - 45x + 672 = 0 \).
This is the required quadratic equation.

Example 5. Ekom is selling tickets for a ride on a giant wheel. If he charges Rs. 30, he will sell 40 tickets. For each Rs. 1 decrease in price, 9 more tickets would get sold. Represent the above problem in the form of a quadratic equation.
Answer:
Let n denote the number of Rs. 1 decrease.
Total revenue can be described as the product of price per ticket and the number of tickets sold: Revenue = price per ticket \( \times \) no. of tickets sold.
The current revenue is \( R = 30 \times 40 \) because there are 40 tickets sold at price of Rs. 30 each.
For each Rs. 1 decrement, the price decreases by Rs. 1, so the new price will be \( (30 - n) \). For each Rs. 1 decrement, 9 more tickets would get sold, so total tickets sold changes from 40 to \( (40 + 9n) \).
New revenue \( = (30 - n)(40 + 9n) = 1200 + 270n - 40n - 9n^2 = -9n^2 + 230n + 1200 \).

Example 6. A fruit grower has 400 crates of apples ready for market and will have 15 more crates each day the shipment is delayed. The price per crate decreases by Rs. 3. The present price is Rs. 70. Form a quadratic equation to represent the above situation.
Answer:
Let n denote the number of days by which the shipment gets delayed. Revenue = price per crate \( \times \) no. of crates sold.
The current revenue is \( R = 70 \times 400 \).
For 1 day delay in the shipment, the price per crate decreases by Rs. 3, so the new price will be \( (70 - 3n) \). For each day delayed, the grower needs to ship 15 more crates, so the number of crates will be \( (400 + 15n) \).
New revenue \( = (70 - 3n)(400 + 15n) = 28000 + 1050n - 1200n - 45n^2 = -45n^2 - 150n + 28000 \).

Solution of Quadratic Equations

The values of x which satisfy the given quadratic equation are known as its roots or zeros or solutions. The solutions of quadratic equations can be found using the following methods: 1. Factorisation method, 2. Completing the square, 3. Quadratic formula.

Solution of Pure Quadratic Equations

Solutions of pure quadratic equations can be found by reducing it to the form \( x^2 = k \) and then the values of x can be determined by taking the square root on both sides. The roots evidently will be \( +\sqrt{k} \) or \( -\sqrt{k} \) written as \( \pm\sqrt{k} \).

Example 7. Solve the given quadratic equations: (i) \( 9x^2 - 28 = 100 + 7x^2 \); (ii) \( (x + 6)(x - 6) = 13 \).
Answer:
(i) Given, \( 9x^2 - 28 = 100 + 7x^2 \Rightarrow 9x^2 - 7x^2 = 128 \Rightarrow x^2 = \frac{128}{2} = 64 \). This is of the form \( x^2 = k \), i.e., this is a pure quadratic equation. \( \therefore x = \pm\sqrt{64} \Rightarrow x = \pm 8 \).
(ii) \( x^2 - 36 = 13 \Rightarrow x^2 = 49 \). This is of the form \( x^2 = k \), i.e., this is a pure quadratic equation. \( \therefore x = \pm\sqrt{49} \Rightarrow x = \pm 7 \).

Example 8. Solve the given quadratic equation for x: \( \frac{x + 2}{x - 2} + \frac{x - 2}{x + 2} = \frac{5}{2} \).
Answer:
\( \frac{x + 2}{x - 2} + \frac{x - 2}{x + 2} = \frac{5}{2} \). Taking the L.C.M.: \( 2\left[(x + 2)^2 + (x - 2)^2\right] = 5(x - 2)(x + 2) \).
\( 2(x + 2)^2 + 2(x - 2)^2 = 5(x^2 - 4) \) [since \( (a - b)(a + b) = a^2 - b^2 \)].
\( 2(x^2 + 4x + 4) + 2(x^2 - 4x + 4) = 5(x^2 - 4) \Rightarrow 2x^2 + 8x + 8 + 2x^2 - 8x + 8 = 5x^2 - 20 \Rightarrow 4x^2 + 16 = 5x^2 - 20 \).
\( x^2 = 36 \Rightarrow x = \pm 6 \).

Example 9. Solve for x: \( (x + 3)(x + 4) + (x - 2)(x - 5) = 30 \).
Answer:
\( x^2 + 7x + 12 + x^2 - 7x + 10 = 30 \Rightarrow 2x^2 + 22 = 30 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2 \).

Solution of Adfected Quadratic Equations

1. Factorisation method

Factorisation refers to the breaking up of an expression (polynomial expression) into smaller expressions. For e.g. \( x^2 - 10x + 21 \) is equal to \( (x - 3)(x - 7) \). Here, a trinomial expression has been broken into binomial expressions.

Let us see some more examples:
1. \( 7x + 42 \) is equal to \( 7(x + 6) \). Here, a binomial expression has been broken into a monomial and a binomial expression.
2. \( x^3 + x^2 - x - 1 \Rightarrow x^2(x + 1) - 1(x + 1) \Rightarrow (x^2 - 1)(x + 1) \). Here, the given polynomial expression has been broken into binomial expressions.

But the most used method in factoring the quadratic equations is known as splitting the middle term.

Algorithm:
Step (i) Let the quadratic equation be \( ax^2 + bx + c,\ a \neq 0 \), i.e., \( 2x^2 - 5x + 3 = 0 \). Here \( a = 2,\ b = -5 \) and \( c = 3 \).
Step (ii) Find the factors of the product of a and c. Here the product of a and c is 6. So, the factors of 6 are \( (1, 6), (2, 3), (3, 2), (6, 1), (-1, -6), (-2, -3), (-3, -2) \) and \( (-6, -1) \).
Step (iii) Express the coefficient of the middle term as the sum or difference of the factors obtained in step (ii). By splitting the middle term, we get \( -5 = (-3) + (-2) \) or \( -5 = (-2) + (-3) \).
Step (iv) Split the middle term in two parts obtained in step (iii). Now, we get \( 2x^2 + [(-2) + (-3)]x + 3 = 0 \Rightarrow 2x^2 - 2x - 3x + 3 = 0 \).
Step (v) Then group the terms to form pairs - the first two terms and the last two terms. Factor each pair by taking out the common factors, i.e., \( 2x(x - 1) - 3(x - 1) = 0 \Rightarrow (2x - 3)(x - 1) = 0 \).
\( \therefore x = 1 \) and \( x = \frac{3}{2} \) are the roots of the given quadratic equation.

Example 10. Solve the quadratic equation: \( x^2 - 3x - 10 = 0 \).
Answer:
The given quadratic equation is \( x^2 - 3x - 10 = 0 \). By comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1,\ b = -3 \) and \( c = -10 \). Here, the product of a and c is \( -10 \).
List down the factors of \( -10 \): \( 1 \times (-10),\ 10 \times (-1),\ 2 \times (-5),\ 5 \times (-2) \). Middle term, \( -3 = (-5) + 2 \).
So, we can split the middle term as follows: \( x^2 - 3x - 10 = x^2 - 5x + 2x - 10 = x(x - 5) + 2(x - 5) = (x + 2)(x - 5) \).
Equating each factor to zero gives \( x + 2 = 0 \Rightarrow x = -2 \) and \( x - 5 = 0 \Rightarrow x = 5 \). Therefore, the solution are \( x = -2 \) and \( x = 5 \).

Example 11. Solve the quadratic equation: \( x^2 + 10x + 25 = 0 \).
Answer:
The given quadratic equation is \( x^2 + 10x + 25 = 0 \). By comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1,\ b = 10,\ c = 25 \). Here, the product of a and c is 25.
List down the factors of 25: \( (1, 25), (5, 5), (25, 1), (-1, -25), (-5, -5) \) and \( (-25, -1) \). Middle term, \( 10 = 5 + 5 \).
So, \( x^2 + 10x + 25 = x^2 + 5x + 5x + 25 = x(x + 5) + 5(x + 5) = (x + 5)(x + 5) \). Therefore, \( x = -5 \) is the answer (a repeated root).

Example 12. Solve \( x^2 + 4x - 5 = 0 \).
Answer:
\( x^2 + 4x - 5 = 0 \Rightarrow x^2 + 5x - x - 5 = 0 \Rightarrow x(x + 5) - 1(x + 5) = 0 \Rightarrow (x - 1)(x + 5) = 0 \).
Therefore, \( x = 1 \) and \( x = -5 \) are the solutions.

Example 13. Factorize: \( 2x^2 + 24x + 64 \).
Answer:
\( 2x^2 + 24x + 64 = 0 \Rightarrow 2x^2 + 16x + 8x + 64 = 0 \Rightarrow 2x(x + 8) + 8(x + 8) = 0 \Rightarrow (2x + 8)(x + 8) = 0 \).
Therefore, \( x = -4 \) and \( x = -8 \) are the solutions.

Example 14. Factorize: \( 2x^2 - 7x + 6 = 3 \).
Answer:
\( 2x^2 - 7x + 6 - 3 = 0 \Rightarrow 2x^2 - 7x + 3 = 0 \Rightarrow 2x^2 - 6x - x + 3 = 0 \Rightarrow 2x(x - 3) - 1(x - 3) = 0 \Rightarrow (2x - 1)(x - 3) = 0 \).
Therefore, \( x = \frac{1}{2} \) and \( x = 3 \) are the solutions of the given quadratic equation.

Example 15. Factorize: \( x^2 + (4 - 3y)x - 12y = 0 \).
Answer:
The given equation is quadratic in x as the highest power of x is 2. Take y as constant.
Expand the equation by opening brackets: \( x^2 + 4x - 3xy - 12y = 0 \).
Factorize: \( x(x + 4) - 3y(x + 4) = 0 \Rightarrow (x + 4)(x - 3y) = 0 \Rightarrow x + 4 = 0 \) or \( x - 3y = 0 \Rightarrow x = -4 \) or \( x = 3y \).
Thus, \( x = -4 \) or \( x = 3y \) are the solutions of the given quadratic equation.