CBSE Class 10 Mathematics Quadratic Equations Notes Set 07

Relation Between Roots and Coefficients

(i) The solutions of the quadratic equation \( ax^2 + bx + c = 0 \) are given by \( x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
(ii) The expression \( b^2 - 4ac = D \) is called the discriminant of the quadratic equation. If \( \alpha, \beta \) are the roots of \( ax^2 + bx + c = 0 \), then

\[ \text{(a) Sum of the roots: } \alpha + \beta = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{b}{a} \]

\[ \text{(b) Product of the roots: } \alpha\beta = \frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{c}{a} \]

(iii) A quadratic equation whose roots are \( \alpha \) and \( \beta \) is \( (x - \alpha)(x - \beta) = 0 \), that is \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \).

Solved Examples

Example 19. Find the quadratic equation having roots 3 and -4.
Answer:
The equation is \( x^2 - (\text{sum of roots})x + \text{product of roots} = 0 \). Sum = 3 - 4 = -1, product = 3(-4) = -12.
So the equation is \( x^2 - (3 - 4)x + 3(-4) = 0 \), that is \( x^2 + x - 12 = 0 \).

Example 20. If \( \alpha, \beta \) are the roots of the equation \( 4x^2 + 3x + 7 = 0 \), find the value of \( \dfrac{1}{\alpha} + \dfrac{1}{\beta} \).
Answer:
\( \alpha + \beta = -\dfrac{3}{4} \) and \( \alpha\beta = \dfrac{7}{4} \).
\( \dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{-3/4}{7/4} = -\dfrac{3}{7} \).

Example 21. The quadratic equation \( ax^2 + bx + c = 0 \) has real roots \( \alpha \) and \( \beta \). If a, b, c are real and of the same sign, determine the sign of its roots.
Answer:
Case I: a, b, c all positive. \( \alpha + \beta = -\dfrac{b}{a} \) is a negative number and \( \alpha\beta = \dfrac{c}{a} \) is a positive number. Two numbers whose sum is negative and product is positive must both be negative.
Case II: a, b, c all negative. Again \( \alpha + \beta = -\dfrac{b}{a} \) is negative and \( \alpha\beta = \dfrac{c}{a} \) is positive, so both roots are negative.
Hence if a, b, c are real and of the same sign, both roots (when real) are negative.

Check Your Progress

1. Find the sum of the roots of the equation \( 2x^2 + 5x + 3 = 0 \).
Answer:
Sum = \( -\dfrac{5}{2} \).

2. Find the value of a if the roots of \( x^2 + (2a + 3)x + 7 = 0 \) are equal in magnitude but opposite in sign.
Answer:
Sum of roots = 0, so \( -(2a + 3) = 0 \), hence \( a = -\dfrac{3}{2} \).

3. Find the value of a if the roots of \( (a + 1)x^2 + 2x + 3a + 7 = 0 \) are reciprocals of each other.
Answer:
Product of roots = 1: \( \dfrac{3a + 7}{a + 1} = 1 \Rightarrow 3a + 7 = a + 1 \Rightarrow a = -3 \).

4. Form a quadratic equation whose roots are 3, -1.
Answer:
\( x^2 - (3 - 1)x + (3)(-1) = 0 \), that is \( x^2 - 2x - 3 = 0 \).

5. If \( \alpha, \beta \) are the roots of \( 3x^2 - 5x + 7 = 0 \), find the value of \( \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} \).
Answer:
\( \alpha + \beta = \dfrac{5}{3} \), \( \alpha\beta = \dfrac{7}{3} \). \( \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} = \dfrac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = \dfrac{25/9 - 14/3}{7/3} = \dfrac{-17/9}{7/3} = -\dfrac{17}{21} \).

Nature of Roots

Consider the quadratic equation \( ax^2 + bx + c = 0 \) having \( \alpha \) and \( \beta \) as its roots. The expression \( b^2 - 4ac \) is called the discriminant of the roots of the quadratic equation and is denoted by D. The roots may be (i) real and unequal, (ii) real and equal, (iii) unreal. If \( a \ne 0 \), b, c are real numbers, the roots are \( \alpha = \dfrac{-b + \sqrt{b^2 - 4ac}}{2a} \) and \( \beta = \dfrac{-b - \sqrt{b^2 - 4ac}}{2a} \). The nature of the roots depends upon the value of the expression \( b^2 - 4ac \) under the square root sign.

Case 1. When \( b^2 - 4ac > 0 \) (D > 0). The roots are real and distinct.
(i) When \( a \ne 0 \), b, c \( \in \) Q and \( b^2 - 4ac \) is a perfect square, both roots are rational and distinct.
(ii) When \( a \ne 0 \), b, c \( \in \) Q and \( b^2 - 4ac \) is not a perfect square, both roots are irrational and distinct.

Case 2. When \( b^2 - 4ac = 0 \) (D = 0). Both roots are real and equal, each equal to \( -\dfrac{b}{2a} \).

Case 3. When \( b^2 - 4ac < 0 \) (D < 0). Then \( \sqrt{b^2 - 4ac} \) is not a real number, so the equation has no real roots (the roots are imaginary and distinct).

Note. If a, b, c \( \in \) Q and \( b^2 - 4ac \) is positive but not a perfect square, then the roots are irrational and always occur in conjugate pairs like \( 2 + \sqrt{3} \) and \( 2 - \sqrt{3} \). However, if a, b, c are irrational numbers and \( b^2 - 4ac \) is positive but not a perfect square, the roots may not occur in conjugate pairs.

Solved Examples

Example 22. Find the nature of the roots of the quadratic equation \( 4x^2 - 5x + 3 = 0 \).
Answer:
Here a = 4, b = -5, c = 3. \( D = b^2 - 4ac = (-5)^2 - 4(4)(3) = 25 - 48 = -23 < 0 \).
Hence the given equation has no real roots.

Example 23. For what value of k is \( (4 - k)x^2 + (2k + 4)x + (8k + 1) = 0 \) a perfect square?
Answer:
The expression is a perfect square if its discriminant is zero.
\( (2k + 4)^2 - 4(4 - k)(8k + 1) = 0 \Rightarrow 4(k + 2)^2 - 4(4 - k)(8k + 1) = 0 \Rightarrow (k^2 + 4k + 4) - (-8k^2 + 31k + 4) = 0 \Rightarrow 9k^2 - 27k = 0 \Rightarrow 9k(k - 3) = 0\).
Hence the given expression is a perfect square if k = 0 or k = 3.

Example 24. If the roots of the equation \( a(b - c)x^2 + b(c - a)x + c(a - b) = 0 \) are equal, show that \( \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c} \).
Answer:
Since the roots are equal, the discriminant is zero: \( b^2(c - a)^2 - 4a(b - c)c(a - b) = 0 \).
\( \Rightarrow b^2(c^2 + a^2 - 2ac) - 4ac(ab - b^2 - ac + bc) = 0 \)
\( \Rightarrow a^2b^2 + b^2c^2 + 4a^2c^2 + 2ab^2c - 4a^2bc - 4abc^2 = 0 \)
\( \Rightarrow (ab + bc - 2ac)^2 = 0 \Rightarrow ab + bc = 2ac \).
Dividing by abc: \( \dfrac{1}{c} + \dfrac{1}{a} = \dfrac{2}{b} \), that is \( \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c} \).

Example 25. If the roots of the equation \( (b - c)x^2 + (c - a)x + (a - b) = 0 \) are equal, prove that 2b = a + c.
Answer:
The discriminant is zero: \( (c - a)^2 - 4(b - c)(a - b) = 0 \).
Expanding: \( (c - a)^2 - 4(ab - b^2 - ac + bc) = c^2 + a^2 - 2ac - 4ab + 4b^2 + 4ac - 4bc = a^2 + c^2 + 4b^2 + 2ac - 4ab - 4bc = 0 \).
\( \Rightarrow (a + c - 2b)^2 = 0 \Rightarrow a + c = 2b \).

Check Your Progress

1. Discuss the nature of the roots of \( 7x^2 - 11x + 4 = 0 \).
Answer:
\( D = 121 - 112 = 9 > 0 \). Real and unequal (rational, as 9 is a perfect square).

2. Discuss the nature of the roots of \( 6x^2 - 13x - 15 = 0 \).
Answer:
\( D = 169 + 360 = 529 = 23^2 > 0 \). Real and unequal (rational).

3. Discuss the nature of the roots of \( x^2 - 7x + 8 = 0 \).
Answer:
\( D = 49 - 32 = 17 > 0 \). Real and unequal (irrational).

4. For what value of k will \( 18x^2 - kx + 2 = 0 \) have equal roots?
Answer:
\( D = k^2 - 144 = 0 \Rightarrow k = \pm 12 \).

5. Show that \( 3x^2 - 7x + 8 = 0 \) cannot be satisfied by any real value of x.
Answer:
\( D = 49 - 96 = -47 < 0 \), so there is no real root.

Word Problems

The method of problem solving consists of three steps. Step (i): translate the word problem into symbolic language (a mathematical statement), that is identify the relationships existing in the problem and form the quadratic equation. Step (ii): solve the quadratic equation thus formed. Step (iii): interpret the solution of the equation, that is translate the result of the mathematical statement into verbal language.

Solved Examples

Example 26. The length of a hall is 5 m more than its breadth. If the area of the floor of the hall is 84 m², what are the length and the breadth of the hall?
Answer:
Let the breadth be x metres; the length is (x + 5) metres. Area: \( x(x + 5) = 84 \Rightarrow x^2 + 5x - 84 = 0 \Rightarrow (x + 12)(x - 7) = 0 \).
So x = 7 or x = -12. The breadth cannot be negative, so we reject x = -12 and take x = 7.
Breadth = 7 m and length = 12 m.

Example 27. Out of a group of swans, \( \dfrac{7}{2} \) times the square root of the total number are playing on the shore of a tank. The two remaining ones are playing in deep water. What is the total number of swans?
Answer:
Let the number of swans be x. Then \( x = \dfrac{7}{2}\sqrt{x} + 2 \), so \( x - 2 = \dfrac{7}{2}\sqrt{x} \).
Squaring: \( 4(x^2 - 4x + 4) = 49x \Rightarrow 4x^2 - 65x + 16 = 0 \Rightarrow 4x(x - 16) - 1(x - 16) = 0 \Rightarrow (x - 16)(4x - 1) = 0 \).
So x = 16 or \( x = \dfrac{1}{4} \). We reject \( x = \dfrac{1}{4} \) (not a whole number) and take x = 16. (Check: \( \dfrac{7}{2}\sqrt{16} = 14 \) and 14 + 2 = 16.)
The total number of swans is 16.

Example 28. The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides is 5 cm. Find the lengths of these sides.
Answer:
Let the shorter side be x cm; the longer side is (x + 5) cm. By the Pythagoras Theorem, \( x^2 + (x + 5)^2 = 25^2 \).
\( 2x^2 + 10x + 25 = 625 \Rightarrow x^2 + 5x - 300 = 0 \Rightarrow (x + 20)(x - 15) = 0 \). We reject x = -20 and take x = 15.
The shorter side is 15 cm and the longer side is 20 cm.

Example 29. Swati can row her boat at a speed of 5 km/h in still water. If it takes her 1 hour more to row the boat 5.25 km upstream than to return downstream, find the speed of the stream.
Answer:
Let the speed of the stream be x km/h. Speed upstream = (5 - x) km/h, speed downstream = (5 + x) km/h.
Time upstream \( t_1 = \dfrac{5.25}{5 - x} \), time downstream \( t_2 = \dfrac{5.25}{5 + x} \). Since \( t_1 = t_2 + 1 \):
\( \dfrac{5.25}{5 - x} = \dfrac{5.25}{5 + x} + 1 \Rightarrow \dfrac{21}{4}\left(\dfrac{1}{5 - x} - \dfrac{1}{5 + x}\right) = 1 \Rightarrow 21 \cdot \dfrac{2x}{25 - x^2} = 4 \Rightarrow 42x = 100 - 4x^2 \).
\( 2x^2 + 21x - 50 = 0 \Rightarrow (2x + 25)(x - 2) = 0 \). We reject \( x = -\dfrac{25}{2} \).
The speed of the stream is 2 km/h.

Example 30. The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the numbers.
Answer:
Let the smaller number be x. The square of the larger number is 18x, so \( x^2 + 18x = 208 \Rightarrow x^2 + 18x - 208 = 0 \Rightarrow (x - 8)(x + 26) = 0 \).
Since the numbers are positive integers, we reject x = -26 and take x = 8. The square of the larger number is 18 × 8 = 144, so the larger number is \( \sqrt{144} = 12 \).
The larger number is 12 and the smaller is 8.

Example 31. The sum S of the first n natural numbers is given by \( S = \dfrac{n(n + 1)}{2} \). Find n, if the sum is 276.
Answer:
\( \dfrac{n(n + 1)}{2} = 276 \Rightarrow n^2 + n - 552 = 0 \). \( n = \dfrac{-1 \pm \sqrt{1 + 2208}}{2} = \dfrac{-1 \pm 47}{2} \), so n = 23 or n = -24.
We reject n = -24, since it is not a natural number. Therefore n = 23.

Check Your Progress

1. The sum of the squares of two consecutive natural numbers is 145. Find the numbers.
Answer:
Let the numbers be n and n + 1: \( n^2 + (n + 1)^2 = 145 \Rightarrow n^2 + n - 72 = 0 \Rightarrow (n + 9)(n - 8) = 0 \). So n = 8.
The numbers are 8, 9.

2. The length of a rectangle is greater than twice its breadth by 2 cm. The length of its diagonal is 13 cm. Find the length and breadth of the rectangle.
Answer:
Breadth b, length 2b + 2: \( b^2 + (2b + 2)^2 = 169 \Rightarrow 5b^2 + 8b - 165 = 0 \Rightarrow (5b + 33)(b - 5) = 0 \). So b = 5.
Length = 12 cm, breadth = 5 cm.

3. The sum of the reciprocals of two consecutive odd natural numbers is \( \dfrac{12}{35} \). Find the numbers.
Answer:
Let the numbers be n and n + 2: \( \dfrac{1}{n} + \dfrac{1}{n + 2} = \dfrac{12}{35} \Rightarrow 35(2n + 2) = 12n(n + 2) \Rightarrow 6n^2 - 23n - 35 = 0 \Rightarrow (n - 5)(6n + 7) = 0\), so n = 5.
The numbers are 5, 7.

4. A car covers a distance of 300 km at a uniform speed. It would cover the same distance in one hour less if its speed were increased by 10 km per hour. Find the speed of the car.
Answer:
\( \dfrac{300}{x} - \dfrac{300}{x + 10} = 1 \Rightarrow x^2 + 10x - 3000 = 0 \Rightarrow (x + 60)(x - 50) = 0 \).
The speed of the car is 50 km/h.

5. A cistern can be filled by two pipes together in \( 33\dfrac{1}{3} \) minutes. The larger pipe takes 15 minutes less than the smaller pipe to fill the cistern separately. Find the time in which each pipe will fill it separately.
Answer:
Let the larger pipe take x minutes and the smaller pipe x + 15 minutes. \( \dfrac{1}{x} + \dfrac{1}{x + 15} = \dfrac{3}{100} \Rightarrow 3x^2 - 155x - 1500 = 0 \Rightarrow (x - 60)(3x + 25) = 0 \), so x = 60.
The larger pipe takes 60 minutes and the smaller pipe takes 75 minutes.