CBSE Class 10 Maths HOTs Quadratic Equations Set 07

Short Answer Type Questions

Question. Check whether the following are quadratic equations or not. (i) \((x - 1)(x + 2) = (x - 3)(x + 1)\) (ii) \((x + 2)^2 = 4(x + 3)\)
Answer: (i) \((x - 1)(x + 2) = (x - 3)(x + 1) \Rightarrow x^2 + x - 2 = x^2 - 2x - 3 \Rightarrow 3x + 1 = 0\). This is not a quadratic equation as the highest power of \(x\) is 1.
(ii) \((x + 2)^2 = 4(x + 3) \Rightarrow x^2 + 4x + 4 = 4x + 12 \Rightarrow x^2 - 8 = 0\). This is a quadratic equation as it is in the form \(ax^2 + bx + c = 0\) where \(a \neq 0\).

Question. If \(x = \frac{1}{\sqrt{3}}\) is root of the equation \(Px^2 + (\sqrt{3} - \sqrt{2})x - 1 = 0\), then find the value of \(P^2 + 1\). 
Answer: Substituting \(x = \frac{1}{\sqrt{3}}\) in \(P(\frac{1}{\sqrt{3}})^2 + (\sqrt{3} - \sqrt{2})(\frac{1}{\sqrt{3}}) - 1 = 0\):
\(\frac{P}{3} + 1 - \frac{\sqrt{2}}{\sqrt{3}} - 1 = 0 \Rightarrow \frac{P}{3} = \frac{\sqrt{2}}{\sqrt{3}} \Rightarrow P = \frac{3\sqrt{2}}{\sqrt{3}} = \sqrt{3} \cdot \sqrt{2} = \sqrt{6}\).
Then \(P^2 + 1 = (\sqrt{6})^2 + 1 = 6 + 1 = 7\).

Question. In each of the following equations, determine the value of \(k\) for which the given value is a solution of the equation. (i) \(kx^2 + 2x - 3 = 0, x = 2\) (ii) \(x^2 + 2ax - k = 0, x = -a\)
Answer: (i) \(k(2)^2 + 2(2) - 3 = 0 \Rightarrow 4k + 1 = 0 \Rightarrow k = -\frac{1}{4}\).
(ii) \((-a)^2 + 2a(-a) - k = 0 \Rightarrow a^2 - 2a^2 - k = 0 \Rightarrow -a^2 - k = 0 \Rightarrow k = -a^2\).

Question. Find the value of \(k\) in the following equations (i) \(x^2 - 2kx - 6 = 0\), when \(x = 3\) (ii) \(x^2 - kx - \frac{5}{4} = 0\), when \(x = \frac{1}{2}\)
Answer: (i) \((3)^2 - 2k(3) - 6 = 0 \Rightarrow 9 - 6k - 6 = 0 \Rightarrow 3 - 6k = 0 \Rightarrow k = \frac{1}{2}\).
(ii) \((\frac{1}{2})^2 - k(\frac{1}{2}) - \frac{5}{4} = 0 \Rightarrow \frac{1}{4} - \frac{k}{2} - \frac{5}{4} = 0 \Rightarrow -1 - \frac{k}{2} = 0 \Rightarrow k = -2\).

Question. Determine whether \(x = -\frac{1}{2}\), \(x = \frac{1}{3}\) are the solutions of the given equation \(6x^2 - x - 2 = 0\), or not.
Answer: For \(x = -\frac{1}{2}\): \(6(-\frac{1}{2})^2 - (-\frac{1}{2}) - 2 = 6(\frac{1}{4}) + \frac{1}{2} - 2 = \frac{3}{2} + \frac{1}{2} - 2 = 2 - 2 = 0\). Yes, it is a solution.
For \(x = \frac{1}{3}\): \(6(\frac{1}{3})^2 - (\frac{1}{3}) - 2 = 6(\frac{1}{9}) - \frac{1}{3} - 2 = \frac{2}{3} - \frac{1}{3} - 2 = \frac{1}{3} - 2 = -\frac{5}{3} \neq 0\). No, it is not a solution.

Question. Solve the quadratic equation by factorisation method. \(4\sqrt{3}x^2 + 5x - 2\sqrt{3} = 0\)
Answer: \(4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3} = 0\)
\(4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) = 0\)
\((4x - \sqrt{3})(\sqrt{3}x + 2) = 0\)
Roots are \(x = \frac{\sqrt{3}}{4}\) and \(x = -\frac{2}{\sqrt{3}}\).

Question. Solve for \(x\): \(\frac{16}{x} - 1 = \frac{15}{x + 1}; x \neq 0, -1\).
Answer: \(\frac{16 - x}{x} = \frac{15}{x + 1} \Rightarrow (16 - x)(x + 1) = 15x\)
\(16x + 16 - x^2 - x = 15x \Rightarrow -x^2 + 15x + 16 = 15x \Rightarrow x^2 = 16\)
\(x = \pm 4\).

Question. Find the roots of the equation \(ax^2 + a^2 = a^2x + x\). 
Answer: \(ax^2 - a^2x - x + a^2 = 0\)
\(a^2x(x/a - 1)\) ... Let's rearrange: \(ax^2 - (a^2 + 1)x + a^2 = 0\)
\(ax^2 - a^2x - x + a^2 = 0 \Rightarrow ax(x - a) - 1(x - a) = 0\)
\((ax - 1)(x - a) = 0\)
Roots are \(x = a, \frac{1}{a}\).

Question. Solve for \(x, \sqrt{6x + 7} - (2x - 7) = 0\) 
Answer: \(\sqrt{6x + 7} = 2x - 7\). Squaring both sides:
\(6x + 7 = (2x - 7)^2 \Rightarrow 6x + 7 = 4x^2 - 28x + 49\)
\(4x^2 - 34x + 42 = 0 \Rightarrow 2x^2 - 17x + 21 = 0\)
\(2x^2 - 14x - 3x + 21 = 0 \Rightarrow 2x(x - 7) - 3(x - 7) = 0\)
\(x = 7\) or \(x = \frac{3}{2}\). Checking solutions, \(x = 7\) works, but \(x = \frac{3}{2}\) makes \(2x - 7\) negative, so \(x = 7\).

Question. Find the numerical difference of the roots of equation \(x^2 - 7x - 18 = 0\). 
Answer: Factorizing \(x^2 - 9x + 2x - 18 = 0 \Rightarrow (x - 9)(x + 2) = 0\). Roots are 9 and \(-2\). Difference = \(9 - (-2) = 11\).

Question. Using the quadratic formula, solve the quadratic equation. \(\sqrt{3}x^2 + 11x + 6\sqrt{3} = 0\)
Answer: \(a = \sqrt{3}, b = 11, c = 6\sqrt{3}\). \(D = 11^2 - 4(\sqrt{3})(6\sqrt{3}) = 121 - 72 = 49\).
\(x = \frac{-11 \pm \sqrt{49}}{2\sqrt{3}} = \frac{-11 \pm 7}{2\sqrt{3}}\).
Roots are \(x = \frac{-4}{2\sqrt{3}} = -\frac{2}{\sqrt{3}}\) and \(x = \frac{-18}{2\sqrt{3}} = -\frac{9}{\sqrt{3}} = -3\sqrt{3}\).

Question. If the discriminant of the equation \(5x^2 - sx + 4 = 0\) is 1, then find the value of \(s\).
Answer: \(D = (-s)^2 - 4(5)(4) = 1\)
\(s^2 - 80 = 1 \Rightarrow s^2 = 81 \Rightarrow s = \pm 9\).

Short Answer Type Questions

Question. Show that \( (x^2 + 1)^2 - x^2 = 0 \) has no real roots. 
Answer: Given that, \( (x^2 + 1)^2 - x^2 = 0 \)
\( \Rightarrow (x^2 + 1)^2 = x^2 \)
\( \Rightarrow x^2 + 1 = \pm x \)
\( \Rightarrow x^2 \mp x + 1 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1, b = \mp 1 \) and \( c = 1 \).
Discriminant, \( D = b^2 - 4ac \)
\( = (\mp 1)^2 - 4 \times 1 \times 1 = 1 - 4 = -3 < 0 \)
Since, discriminant is negative, therefore quadratic equation \( (x^2 + 1)^2 - x^2 = 0 \) has no real roots i.e. imaginary roots.

Question. Find the value of \( k \) for which the quadratic equation \( 2x^2 - kx + k = 0 \) has equal roots. 
Answer: Given equation is \( 2x^2 - kx + k = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get \( a = 2, b = -k \) and \( c = k \).
For equal roots, the discriminant must be zero.
i.e. \( D = b^2 - 4ac = 0 \)
\( \Rightarrow (-k)^2 - 4(2)(k) = 0 \)
\( \Rightarrow k^2 - 8k = 0 \)
\( \Rightarrow k(k - 8) = 0 \)
\( \therefore k = 0, 8 \)
Hence, the required values of \( k \) are 0 and 8.

Question. Find the values of \( k \) for which the equation \( 9x^2 + 3kx + 4 = 0 \) has real roots.
Answer: Given quadratic equation is \( 9x^2 + 3kx + 4 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get \( a = 9, b = 3k \) and \( c = 4 \).
Now, \( D = b^2 - 4ac = (3k)^2 - 4(9)(4) = 9k^2 - 144 \)
Since, roots of given equation are real.
\( \therefore D \geq 0 \Rightarrow 9k^2 - 144 \geq 0 \)
\( \Rightarrow 9(k^2 - 16) \geq 0 \)
\( \Rightarrow k^2 - 16 \geq 0 \)
\( \Rightarrow k^2 - (4)^2 \geq 0 \)
\( \Rightarrow (k - 4)(k + 4) \geq 0 \)
\( \Rightarrow k \leq -4 \) or \( k \geq 4 \)

Question. If the equation \( (1 + m^2)x^2 + (2mc)x + (c^2 - a^2) = 0 \) has equal roots, then prove that \( c^2 = a^2(1 + m^2) \).
Answer: Given equation is \( (1 + m^2)x^2 + (2mc)x + (c^2 - a^2) = 0 \)
On comparing with \( Ax^2 + Bx + C = 0 \), we get \( A = (1 + m^2), B = 2mc \) and \( C = (c^2 - a^2) \).
Since, the given equation has equal roots.
\( \therefore \) Discriminant, \( D = 0 \Rightarrow B^2 - 4AC = 0 \)
\( \Rightarrow (2mc)^2 - 4(1 + m^2)(c^2 - a^2) = 0 \)
\( \Rightarrow 4m^2c^2 - 4(c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \)
\( \Rightarrow m^2c^2 - (c^2 - a^2 + m^2c^2 - m^2a^2) = 0 \) [dividing by 4]
\( \Rightarrow m^2c^2 - c^2 + a^2 - m^2c^2 + m^2a^2 = 0 \)
\( \Rightarrow -c^2 + a^2 + m^2a^2 = 0 \)
\( \Rightarrow -c^2 + a^2(1 + m^2) = 0 \)
\( \Rightarrow c^2 = a^2(1 + m^2) \). Hence proved.

Question. The sum of two numbers is 11 and the sum of their reciprocals is \( \frac{11}{28} \). Find the numbers. 
Answer: Let one number be \( x \). Then, another number = \( (11 - x) \).
According to the question, \( \frac{1}{x} + \frac{1}{11 - x} = \frac{11}{28} \)
\( \Rightarrow \frac{11 - x + x}{x(11 - x)} = \frac{11}{28} \)
\( \Rightarrow \frac{11}{x(11 - x)} = \frac{11}{28} \)
\( \Rightarrow x(11 - x) = 28 \)
\( \Rightarrow 11x - x^2 = 28 \Rightarrow x^2 - 11x + 28 = 0 \)
\( \Rightarrow x^2 - 7x - 4x + 28 = 0 \)
\( \Rightarrow x(x - 7) - 4(x - 7) = 0 \)
\( \Rightarrow (x - 7)(x - 4) = 0 \)
\( \Rightarrow x = 4 \) or \( x = 7 \)
When \( x = 4 \), then \( 11 - x = 11 - 4 = 7 \)
When \( x = 7 \), then \( 11 - x = 11 - 7 = 4 \)
Hence, the numbers are 4 and 7.

Question. In a cricket match, Harbhajan took three wickets less than twice the number of wickets taken by Zaheer. The product of the numbers of wickets taken by these two is 20. Represent the above situation in the form of a quadratic equation. 
Answer: Let the number of wickets taken by Zaheer in a cricket match are \( x \), then number of wickets taken by Harbhajan = \( 2x - 3 \).
According to the question,
\( x(2x - 3) = 20 \)
\( \Rightarrow 2x^2 - 3x = 20 \)
\( \Rightarrow 2x^2 - 3x - 20 = 0 \)

Long Answer Type Questions

Question. If \( x = 2 \) and \( x = 3 \) are roots of the equation \( 3x^2 - 2ax + 2b = 0 \), then find the values of \( a \) and \( b \).
Answer: Given, \( 3x^2 - 2ax + 2b = 0 \)
Since, \( x = 2 \) and \( x = 3 \) are the solutions of given equation, so it will satisfy the given equation.
On putting \( x = 2 \) and \( x = 3 \) one-by-one, we get
\( 3(2)^2 - 2a(2) + 2b = 0 \)
\( \Rightarrow 12 - 4a + 2b = 0 \)
\( \Rightarrow -2(2a - b - 6) = 0 \)
\( \Rightarrow 2a - b = 6 \)
and \( 3(3)^2 - 2a(3) + 2b = 0 \)
\( \Rightarrow 27 - 6a + 2b = 0 \)
\( \Rightarrow 6a - 2b = 27 \)
On multiplying Eq. (ii) by 2 and then subtract it from Eq. (iii), we get
\( 6a - 2b - 4a + 2b = 27 - 12 \)
\( \Rightarrow 2a = 15 \Rightarrow a = \frac{15}{2} \)
On substituting \( a = \frac{15}{2} \) in Eq. (ii), we get
\( 2 \times \frac{15}{2} - b = 6 \)
\( \Rightarrow 15 - b = 6 \Rightarrow b = 15 - 6 = 9 \)
Hence, the required values of \( a \) and \( b \) are \( 15/2 \) and 9, respectively.

Question. Find the nature of roots of the following quadratic equations. If the real roots exist, then also find the roots.
(i) \( 4x^2 + 12x + 9 = 0 \)
(ii) \( 3x^2 + 5x - 7 = 0 \)
Answer: (i) Given quadratic equation is \( 4x^2 + 12x + 9 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get \( a = 4, b = 12 \) and \( c = 9 \).
Now, \( D = b^2 - 4ac = (12)^2 - 4(4)(9) = 144 - 144 = 0 \)
Since, \( D = 0 \), so given quadratic equation has two equal and real roots which are given by
\( x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-12 \pm 0}{2(4)} \)
\( \Rightarrow x = \frac{-12 + 0}{8} \) or \( x = \frac{-12 - 0}{8} \)
\( \Rightarrow x = -\frac{3}{2} \) or \( x = -\frac{3}{2} \)
Hence, the roots are \( -\frac{3}{2} \) and \( -\frac{3}{2} \).

(ii) Given quadratic equation is \( 3x^2 + 5x - 7 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get \( a = 3, b = 5 \) and \( c = -7 \).
Now, \( D = b^2 - 4ac = (5)^2 - 4(3)(-7) = 25 + 84 = 109 \)
Since, \( D > 0 \), so given quadratic equation has two distinct real roots which are given by
\( x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-5 \pm \sqrt{109}}{2(3)} \)
\( \Rightarrow x = \frac{-5 + \sqrt{109}}{6} \) or \( x = \frac{-5 - \sqrt{109}}{6} \)
Hence, the roots are \( \frac{-5 + \sqrt{109}}{6} \) and \( \frac{-5 - \sqrt{109}}{6} \).

Question. Find the value of \( k \) for which the given equation has equal roots.
\( (k - 12)x^2 + 2(k - 12)x + 2 = 0 \)
Answer: Given quadratic equation is \( (k - 12)x^2 + 2(k - 12)x + 2 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get \( a = (k - 12), b = 2(k - 12) \) and \( c = 2 \).
Now, \( D = b^2 - 4ac = [2(k - 12)]^2 - 4(k - 12)(2) \)
\( = 4(k - 12)^2 - 8(k - 12) = 4(k - 12)[(k - 12) - 2] = 4(k - 12)(k - 14) \)
Since, roots of given equation are equal, \( \therefore D = 0 \)
\( \Rightarrow 4(k - 12)(k - 14) = 0 \)
\( \Rightarrow k - 12 = 0 \) or \( k - 14 = 0 \)
\( \Rightarrow k = 12 \) or \( k = 14 \)
But \( k = 12 \) does not satisfy the given equation because if \( k = 12 \), then coefficients of \( x^2 \) and \( x \) become zero.
Hence, required value of \( k \) is 14.

Question. If \( x = -2 \) is a root of the equation \( 3x^2 + 7x + p = 0 \). Find the values of \( k \), so that the roots of the equation \( x^2 + k(4x + k - 1) + p = 0 \) are equal. 
Answer: Given equation is \( 3x^2 + 7x + p = 0 \)
Since, \( x = -2 \) is a root of the given equation, so it will satisfy the given equation.
On putting \( x = -2 \) in the given equation, we get
\( 3(-2)^2 + 7(-2) + p = 0 \Rightarrow 12 - 14 + p = 0 \Rightarrow p = 2 \)
On putting \( p = 2 \) in \( x^2 + k(4x + k - 1) + p = 0 \), we get
\( x^2 + 4kx + k^2 - k + 2 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1, b = 4k \) and \( c = k^2 - k + 2 \).
\( \therefore D = b^2 - 4ac = (4k)^2 - 4(1)(k^2 - k + 2) = 16k^2 - 4k^2 + 4k - 8 = 12k^2 + 4k - 8 \)
Since, roots are equal, \( \therefore D = 0 \)
\( \Rightarrow 12k^2 + 4k - 8 = 0 \Rightarrow 3k^2 + k - 2 = 0 \)
\( \Rightarrow 3k^2 + 3k - 2k - 2 = 0 \Rightarrow 3k(k + 1) - 2(k + 1) = 0 \)
\( \Rightarrow (3k - 2)(k + 1) = 0 \Rightarrow k = \frac{2}{3}, -1 \)

Question. Find two consecutive odd natural numbers, sum of whose squares is 130. 
Answer: Let two consecutive odd natural numbers are \( x \) and \( x + 2 \).
Then according to the given condition, \( x^2 + (x + 2)^2 = 130 \)
\( \Rightarrow x^2 + x^2 + 4x + 4 = 130 \Rightarrow 2x^2 + 4x - 126 = 0 \)
\( \Rightarrow x^2 + 2x - 63 = 0 \Rightarrow x^2 + 9x - 7x - 63 = 0 \)
\( \Rightarrow x(x + 9) - 7(x + 9) = 0 \Rightarrow (x - 7)(x + 9) = 0 \)
\( \Rightarrow x = 7, -9 \)
Since, natural number cannot be negative, so we neglect \( x = -9 \).
Thus, \( x = 7 \) and \( x + 2 = 7 + 2 = 9 \).
Hence, two consecutive odd numbers are 7 and 9.

Question. A piece of cloth costs ₹ 200. If the piece was 5 m longer and each metre of cloth costs ₹ 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre? 
Answer: Let the length of piece be \( x \) m. Then, \( \text{rate} = \frac{200}{x} \text{ per m} \).
Now, \( \text{new length} = (x + 5) \text{ m} \). Since, the cost remains same, \( \therefore \text{New rate} = \frac{200}{x + 5} \text{ per m} \).
According to the given condition, \( \frac{200}{x} - \frac{200}{x + 5} = 2 \)
\( \Rightarrow \frac{100}{x} - \frac{100}{x + 5} = 1 \Rightarrow 100(x + 5) - 100x = x(x + 5) \)
\( \Rightarrow 100x + 500 - 100x = x^2 + 5x \Rightarrow x^2 + 5x - 500 = 0 \)
\( \Rightarrow x^2 + 25x - 20x - 500 = 0 \Rightarrow (x + 25)(x - 20) = 0 \Rightarrow x = 20, -25 \).
Since length cannot be negative, neglect \( x = -25 \).
Thus, \( x = 20 \). Now, \( \text{rate} = \frac{200}{20} = Rs 10 \).
Hence, length of piece is 20 m and rate per metre is Rs 10.

Question. The difference of two numbers is 4. If the difference of their reciprocals is \( \frac{4}{21} \), find the two numbers. 
Answer: Let first number be \( x \). Then, second number = \( x + 4 \).
According to the question, \( \frac{1}{x} - \frac{1}{x + 4} = \frac{4}{21} \)
\( \Rightarrow \frac{x + 4 - x}{x(x + 4)} = \frac{4}{21} \Rightarrow \frac{4}{x^2 + 4x} = \frac{4}{21} \Rightarrow x^2 + 4x = 21 \)
\( \Rightarrow x^2 + 4x - 21 = 0 \Rightarrow x^2 + 7x - 3x - 21 = 0 \)
\( \Rightarrow x(x + 7) - 3(x + 7) = 0 \Rightarrow (x - 3)(x + 7) = 0 \)
\( \Rightarrow x = 3, -7 \).
When \( x = 3 \), second number \( = 3 + 4 = 7 \).
When \( x = -7 \), second number \( = -7 + 4 = -3 \).
Hence, two numbers are 3, 7 or -7, -3.

Question. The perimeter of a right angled triangle is 70 units and its hypotenuse is 29 units we would like to find the length of the other sides.
Answer: Let one side = \( x \). Now, perimeter of a triangle, \( 70 = x + 29 + AB \)
\( \Rightarrow AB = 70 - 29 - x = 41 - x \).
In right \( \Delta ABC \), use Pythagoras theorem, \( BC^2 = AC^2 + AB^2 \)
\( \Rightarrow (29)^2 = x^2 + (41 - x)^2 \Rightarrow 841 = x^2 + 1681 + x^2 - 82x \)
\( \Rightarrow 2x^2 - 82x + 840 = 0 \Rightarrow x^2 - 41x + 420 = 0 \)
\( \Rightarrow x^2 - 21x - 20x + 420 = 0 \Rightarrow x(x - 21) - 20(x - 21) = 0 \)
\( \Rightarrow (x - 20)(x - 21) = 0 \Rightarrow x = 20, 21 \).
Hence, length of other sides of a \( \Delta ABC \) are 20 units, 21 units.

Question. The sum of the reciprocals of Anjali’s age 3 yr ago and 5 yr from now is \( \frac{1}{3} \). Find the present age of Anjali.
Answer: Let present age of Anjali be \( x \) yr.
\( \text{Anjali’s age 3 yr ago} = (x - 3) \text{ yr} \)
\( \text{Anjali’s age 5 yr from now} = (x + 5) \text{ yr} \)
According to the question, \( \frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3} \)
\( \Rightarrow \frac{x + 5 + x - 3}{(x - 3)(x + 5)} = \frac{1}{3} \Rightarrow \frac{2x + 2}{x^2 + 5x - 3x - 15} = \frac{1}{3} \)
\( \Rightarrow 3(2x + 2) = x^2 + 2x - 15 \Rightarrow 6x + 6 = x^2 + 2x - 15 \)
\( \Rightarrow x^2 + 2x - 15 - 6x - 6 = 0 \Rightarrow x^2 - 4x - 21 = 0 \)
Now, by factorisation method, \( x^2 - 7x + 3x - 21 = 0 \)
\( \Rightarrow x(x - 7) + 3(x - 7) = 0 \Rightarrow (x - 7)(x + 3) = 0 \)
\( \Rightarrow x = 7, -3 \).
Age cannot be negative, so \( x = 7 \). Hence, Anjali's present age is 7 yr.

Question. A two-digit number is such that the product of the digits is 12. When 36 is added to the number the digits interchange their places. Find the two-digit number.
Answer: Let the ten's digit of the number be \( x \).
Product of the digits = 12 \( \Rightarrow \text{Unit's digit} = \frac{12}{x} \).
Two-digit number \( = 10x + \frac{12}{x} \).
If 36 is added, digits interchange: \( 10x + \frac{12}{x} + 36 = 10 \left( \frac{12}{x} \right) + x \)
\( \Rightarrow 10x^2 + 12 + 36x = 120 + x^2 \Rightarrow 9x^2 + 36x - 108 = 0 \)
\( \Rightarrow x^2 + 4x - 12 = 0 \Rightarrow (x + 6)(x - 2) = 0 \Rightarrow x = -6, 2 \).
Since a digit can never be negative, \( x = 2 \).
Two-digit number \( = 10(2) + \frac{12}{2} = 20 + 6 = 26 \).

Question. John and Janvi together have 45 marbles. Both of them lost 5 marbles each and the product of the number of marbles they now have, is 124. Find out how many marbles they had to start with?
Answer: Let John has \( x \) marbles. Then, number of marbles Janvi has \( = 45 - x \).
After losing 5 marbles each, John has \( (x - 5) \) and Janvi has \( (45 - x - 5) = (40 - x) \).
Product \( = 124 \Rightarrow (x - 5)(40 - x) = 124 \)
\( \Rightarrow 40x - x^2 - 200 + 5x = 124 \Rightarrow -x^2 + 45x - 324 = 0 \Rightarrow x^2 - 45x + 324 = 0 \)
\( \Rightarrow x^2 - 36x - 9x + 324 = 0 \Rightarrow (x - 36)(x - 9) = 0 \Rightarrow x = 36, 9 \).
If John has 36, Janvi has 9. If John has 9, Janvi has 36.

Question. The hypotenuse of right angled triangle is 6 m more than twice the shortest side. If the third side is 2 m less than the hypotenuse, then find all sides of the triangle. [CBSE 2020 (Standard)]
Answer: Let length of shortest side = \( x \) m. Then, hypotenuse = \( (2x + 6) \) m and third side = \( (2x + 6 - 2) = (2x + 4) \) m.
By Pythagoras theorem, \( (2x + 6)^2 = x^2 + (2x + 4)^2 \)
\( \Rightarrow 4x^2 + 24x + 36 = x^2 + 4x^2 + 16x + 16 \Rightarrow x^2 - 8x - 20 = 0 \)
\( \Rightarrow (x - 10)(x + 2) = 0 \Rightarrow x = 10, -2 \).
Shortest side = 10 m, Hypotenuse = \( 2(10) + 6 = 26 \) m, Third side = \( 2(10) + 4 = 24 \) m.

Question. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age. Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha. [NCERT Exemplar]
Answer: Let Nisha's age = \( x \) yr, then Asha's age = \( (x^2 + 2) \) yr.
Difference in age \( = (x^2 + 2 - x) \). After this many years, Nisha becomes \( (x^2 + 2) \).
Asha's age will be \( (x^2 + 2) + (x^2 + 2 - x) = 2x^2 - x + 4 \).
Condition: \( 2x^2 - x + 4 = 10x - 1 \Rightarrow 2x^2 - 11x + 5 = 0 \)
\( \Rightarrow 2x^2 - 10x - x + 5 = 0 \Rightarrow (2x - 1)(x - 5) = 0 \Rightarrow x = 5 \) (since \( x = 1/2 \) is not possible).
Nisha = 5 yr, Asha = \( 5^2 + 2 = 27 \) yr.

Question. The speed of a boat in still water is 15 km/h. It can go 30 km upstream and return downstream to the original point in 4 h and 30 min. Find the speed of stream.
Answer: Let speed of stream = \( x \) km/h. Speed upstream = \( (15 - x) \), downstream = \( (15 + x) \).
\( \frac{30}{15 - x} + \frac{30}{15 + x} = 4 \frac{1}{2} = \frac{9}{2} \)
\( \Rightarrow 30 \left[ \frac{15 + x + 15 - x}{(15 - x)(15 + x)} \right] = \frac{9}{2} \Rightarrow \frac{30(30)}{225 - x^2} = \frac{9}{2} \)
\( \Rightarrow \frac{900}{225 - x^2} = \frac{9}{2} \Rightarrow 200 = 225 - x^2 \Rightarrow x^2 = 25 \Rightarrow x = 5 \).
Speed of stream = 5 km/h.

Question. Two water taps together can fill a tank in \( 1 \frac{7}{8} \) h. The tap with longer diameter takes 2 h less than the tap with smaller one to fill the tank separately. Find the time in which each tap can fill the tank separately. [CBSE 2019]
Answer: Let smaller tap take \( x \) h. Larger tap takes \( (x - 2) \) h.
In 1 h, together they fill \( \frac{1}{x} + \frac{1}{x - 2} \).
Total time \( = \frac{15}{8} \text{ h} \Rightarrow \frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15} \)
\( \Rightarrow \frac{x - 2 + x}{x(x - 2)} = \frac{8}{15} \Rightarrow 15(2x - 2) = 8(x^2 - 2x) \)
\( \Rightarrow 30x - 30 = 8x^2 - 16x \Rightarrow 8x^2 - 46x + 30 = 0 \Rightarrow 4x^2 - 23x + 15 = 0 \)
\( \Rightarrow (x - 5)(4x - 3) = 0 \Rightarrow x = 5, 3/4 \).
Smaller tap = 5 h, larger tap = 3 h (since \( x = 3/4 \) makes \( x - 2 \) negative).

Case Based Questions

In the centre of a rectangular lawn of dimensions \( 50\text{ m} \times 40\text{ m} \), a rectangular pond has to be constructed, so that the area of the grass surrounding the pond would be \( 1184\text{ m}^2 \).

Question. If the distance between pond and lawn is \( x\text{ m} \). Find the length and breadth of rectangular pond.
Answer: Length of rectangular pond \( (l_2) = 50 - 2x \).
Breadth of rectangular pond \( (b_2) = 40 - 2x \).

Question. Find the quadratic equation related to the given problem.
Answer: Area of grass = Area of lawn - Area of pond
\( 1184 = (50 \times 40) - (50 - 2x)(40 - 2x) \)
\( \Rightarrow 1184 = 2000 - [2000 - 100x - 80x + 4x^2] \)
\( \Rightarrow 1184 = 180x - 4x^2 \Rightarrow 4x^2 - 180x + 1184 = 0 \)
\( \Rightarrow x^2 - 45x + 296 = 0 \).

Question. Find the length and breadth of the pond.
Answer: \( x^2 - 45x + 296 = 0 \Rightarrow (x - 37)(x - 8) = 0 \)
\( \Rightarrow x = 8 \) (since \( x = 37 \) gives negative dimensions).
Length = \( 50 - 2(8) = 34\text{ m} \).
Breadth = \( 40 - 2(8) = 24\text{ m} \).