CBSE Class 10 Mathematics Quadratic Equations VBQs Set 17

Check whether the following are quadratic equations:

Question. \( (x + 1)^2 = 2 (x - 3) \)
Answer: \( (x + 1)^2 = 2 (x - 3) \)
\( \implies \) \( x^2 + 2x + 1 = 2x - 6 \)
\( \implies \) \( x^2 + 2x - 2x + 1 + 6 = 0 \)
\( \implies \) \( x^2 + 7 = 0 \)
or \( x^2 + 0.x + 7 = 0 \)
Which is of the form \( ax^2 + bx + c = 0 \). Hence, the given equation is a quadratic equation.

Question. \( x^2 - 2x = (- 2) (3 - x) \)
Answer: \( x^2 - 2x = (- 2) (3 - x) \)
\( \implies \) \( x^2 - 2x = - 6 + 2x \)
\( \implies \) \( x^2 - 4x + 6 = 0 \)
Which is of the form \( ax^2 + bx + c = 0 \). Hence, the given equation is a quadratic equation.

Question. \( (x - 2) (x + 1) = (x - 1) (x + 3) \)
Answer: \( (x - 2) (x + 1) = (x - 1) (x + 3) \)
\( \implies \) \( x^2 + x - 2x - 2 = x^2 + 3x - x - 3 \)
\( \implies \) \( x^2 + x - 2x - 2 - x^2 - 3x + x + 3 = 0 \)
\( \implies \) \( - 3x + 1 = 0 \)
\( \implies \) \( 3x - 1 = 0 \)
or \( 0.x^2 + 3x + (-1) = 0 \)
Which is of the form \( ax^2 + bx + c = 0 \), But \( a = 0 \). Hence, given equation is not a quadratic equation.

Question. \( (x - 3) (2x + 1) = x (x + 5) \)
Answer: \( (x - 3) (2x + 1) = x (x + 5) \)
\( \implies \) \( 2x^2 + x - 6x - 3 = x^2 + 5x \)
\( \implies \) \( x^2 - 10x - 3 = 0 \)
Which is of the form \( ax^2 + bx + c = 0 \). Hence, the given equation is a quadratic equation.

Question. \( (2x - 1) (x - 3) = (x + 5) (x - 1) \)
Answer: \( (2x - 1) (x - 3) = (x + 5) (x - 1) \)
\( \implies \) \( 2x^2 - 6x - x + 3 = x^2 - x + 5x - 5 \)
\( \implies \) \( x^2 - 11x + 8 = 0 \)
Which is of the form \( ax^2 + bx + c = 0 \). Hence, the given equation is a quadratic equation.

Question. \( x^2 + 3x + 1 = (x - 2)^2 \)
Answer: \( x^2 + 3x + 1 = (x - 2)^2 \)
\( \implies \) \( x^2 + 3x + 1 = x^2 + 4 - 4x \)
\( \implies \) \( x^2 + 3x + 1 - x^2 - 4 + 4x = 0 \)
\( \implies \) \( 0.x^2 + 7x - 3 = 0 \)
Which is of the form \( ax^2 + bx + c = 0 \), but \( a = 0 \). Hence, the given equation is not a quadratic equation.

Question. \( (x + 2)^3 = 2x(x^2 - 1) \)
Answer: \( (x + 2)^3 = 2x(x^2 - 1) \)
\( \implies \) \( x^3 + 8 + 3 \cdot x \cdot 2(x + 2) = 2x^3 - 2x \)
\( \implies \) \( x^3 - 6x^2 - 14x - 8 = 0 \)
Which is not of the form \( ax^2 + bx + c = 0 \). Hence, the given equation is not a quadratic equation.

Question. \( x^3 - 4x^2 - x + 1 = (x - 2)^3 \)
Answer: \( x^3 - 4x^2 - x + 1 = (x - 2)^3 \)
\( \implies \) \( x^3 - 4x^2 - x + 1 = x^3 - 8 + 3 \cdot x(- 2)(x - 2) \)
\( \implies \) \( 2x^2 - 13x + 9 = 0 \)
Which is of the form \( ax^2 + bx + c = 0 \). Hence, the given equation is a quadratic equation.

Represent the following situations in the form of quadratic equations:

Question. The area of a rectangular plot is 528 m\( ^2 \). The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Answer: Let breadth of the rectangular plot be \( x \) m
\( \implies \) Length of the plot = \( (2x + 1) \) m
Area of a rectangular plot = \( l \times b \)
\( \implies \) \( 528 = (2x + 1)x \)
\( \implies \) \( 528 = 2x^2 + x \)
\( \implies \) \( 2x^2 + x - 528 = 0 \)
Which is the required quadratic equation.

Question. The product of two consecutive positive integers is 306. We need to find the integers.
Answer: Let two consecutive integers be \( x \) and \( x + 1 \).
Then, \( x(x + 1) = 306 \)
\( \implies \) \( x^2 + x - 306 = 0 \)
Which is the required quadratic equation.

Question. Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Answer: Let present age of Rohan be \( x \) years
Rohan’s mother’s present age be \( (x + 26) \) years
After 3 years, Rohan’s age = \( (x + 3) \) years
After 3 years, Rohan’s mother’s age = \( (x + 26 + 3) \) years = \( (x + 29) \) years
ATQ \( (x + 3)(x + 29) = 360 \)
\( \implies \) \( x^2 + 32x - 273 = 0 \)
Which is the required quadratic equation.

Question. A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Answer: Let speed of the train be \( x \) km/h
Total distance to be covered = 480 km
Time = \( \frac{\text{distance}}{\text{speed}} = \frac{480}{x} \) hours
Decreased speed of the train = \( (x - 8) \) km/h
Now, time = \( \frac{480}{x - 8} \) hours
ATQ \( \frac{480}{x - 8} - \frac{480}{x} = 3 \)
\( \implies \) \( 480 \left[ \frac{1}{x - 8} - \frac{1}{x} \right] = 3 \)
\( \implies \) \( 480 \left[ \frac{x - x + 8}{x(x - 8)} \right] = 3 \)
\( \implies \) \( 480 \times 8 = 3x (x - 8) \)
\( \implies \) \( 3840 = 3x^2 - 24x \)
\( \implies \) \( 3x^2 - 24x - 3840 = 0 \)
or \( x^2 - 8x - 1280 = 0 \)
Which is the required quadratic equation.

Find the roots of the following quadratic equations by factorisation:

Question. \( x^2 - 3x - 10 = 0 \) 
Answer: \( x^2 - 3x - 10 = 0 \)
\( \implies \) \( x^2 - 5x + 2x - 10 = 0 \)
\( \implies \) \( x(x - 5) + 2(x - 5) = 0 \)
\( \implies \) \( (x + 2)(x - 5) = 0 \)
\( \implies \) \( x + 2 = 0 \) or \( x - 5 = 0 \)
\( \implies \) \( x = -2 \) or \( x = 5 \)
Hence, the roots are 5, -2.

Question. \( 2x^2 + x - 6 = 0 \)
Answer: \( 2x^2 + x - 6 = 0 \)
\( \implies \) \( 2x^2 + 4x - 3x - 6 = 0 \)
\( \implies \) \( 2x(x + 2) - 3(x + 2) = 0 \)
\( \implies \) \( (2x - 3)(x + 2) = 0 \)
\( \implies \) \( 2x - 3 = 0 \) or \( x + 2 = 0 \)
\( \implies \) \( 2x = 3 \) or \( x = -2 \)
\( \implies \) \( x = \frac{3}{2} \) or \( x = -2 \)
Hence, the roots are -2, \( \frac{3}{2} \).

Question. \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
Answer: \( \sqrt{2}x^2 + 7x + 5\sqrt{2} = 0 \)
\( \implies \) \( \sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0 \)
\( \implies \) \( x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0 \)
\( \implies \) \( (x + \sqrt{2})(\sqrt{2}x + 5) = 0 \)
\( \implies \) \( x + \sqrt{2} = 0 \) or \( \sqrt{2}x + 5 = 0 \)
\( \implies \) \( x = -\sqrt{2} \) or \( x = -\frac{5}{\sqrt{2}} \)
Hence, the roots are \( -\sqrt{2}, -\frac{5}{\sqrt{2}} \).

Question. \( 2x^2 - x + \frac{1}{8} = 0 \)
Answer: \( \frac{16x^2 - 8x + 1}{8} = 0 \)
\( \implies \) \( 16x^2 - 4x - 4x + 1 = 0 \)
\( \implies \) \( 4x(4x - 1) - 1(4x - 1) = 0 \)
\( \implies \) \( (4x - 1)(4x - 1) = 0 \)
\( \implies \) \( 4x - 1 = 0 \)
or \( 4x - 1 = 0 \)
\( \implies \) \( x = \frac{1}{4} \)
or \( x = \frac{1}{4} \)
Hence, the roots are \( \frac{1}{4}, \frac{1}{4} \).

Question. \( 100x^2 - 20x + 1 = 0 \)
Answer: \( 100x^2 - 10x - 10x + 1 = 0 \)
\( \implies \) \( 10x(10x - 1) - 1(10x - 1) = 0 \)
\( \implies \) \( (10x - 1)(10x - 1) = 0 \)
\( \implies \) \( 10x - 1 = 0 \) or \( 10x - 1 = 0 \)
\( \implies \) \( x = \frac{1}{10} \) or \( x = \frac{1}{10} \)
Hence, the roots are \( \frac{1}{10}, \frac{1}{10} \).

Solve the following situations mathematically:

Question. 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 would like to find out how many marbles they had to start with.
Answer: Let the number of marbles John had be \( x \)
Then the number of marbles Jivanti had be \( 45 - x \)
The number of marbles left with John, when he lost 5 marbles = \( x - 5 \)
The number of marbles left with Jivanti, when she lost 5 marbles = \( 45 - x - 5 = 40 - x \)
ATQ \( (x - 5)(40 - x) = 124 \)
\( \implies \) \( 40x - x^2 - 200 + 5x = 124 \)
\( \implies \) \( x^2 - 45x + 324 = 0 \)
\( \implies \) \( x^2 - 36x - 9x + 324 = 0 \)
\( \implies \) \( x(x - 36) - 9(x - 36) = 0 \)
\( \implies \) \( (x - 9)(x - 36) = 0 \)
\( \implies \) \( x - 9 = 0 \) or \( x - 36 = 0 \)
\( \implies \) \( x = 9 \) or \( x = 36 \)
if \( x = 9 \), then \( 45 - x = 45 - 9 = 36 \)
if \( x = 36 \), then \( 45 - x = 45 - 36 = 9 \)
Hence, number of marbles they had to start with 9 and 36.

Question. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was Rs. 750. We would like to find out the number of toys produced on that day.
Answer: Let the number of toys produced in a day be \( x \)
Then, cost of production of each toy on that day be Rs. \( (55 - x) \)
Total cost of production on that day = Rs. \( x(55 - x) \)
ATQ \( x(55 - x) = 750 \)
\( \implies \) \( 55x - x^2 = 750 \)
\( \implies \) \( x^2 - 55x + 750 = 0 \)
\( \implies \) \( x^2 - 25x - 30x + 750 = 0 \)
\( \implies \) \( (x - 30)(x - 25) = 0 \)
\( \implies \) \( x - 30 = 0 \) or \( x - 25 = 0 \)
\( \implies \) \( x = 30 \) or \( x = 25 \)
Number of toys produced on that day was 25 or 30.

Question. Find two numbers whose sum is 27 and product is 182.
Answer: Let one number be \( x \), then other number be \( 27 - x \)
ATQ \( x(27 - x) = 182 \)
\( \implies \) \( x^2 - 27x + 182 = 0 \)
\( \implies \) \( x^2 - 14x - 13x + 182 = 0 \)
\( \implies \) \( x(x - 14) - 13(x - 14) = 0 \)
\( \implies \) \( (x - 13)(x - 14) = 0 \)
\( \implies \) \( x - 13 = 0 \) or \( x - 14 = 0 \)
\( \implies \) \( x = 13 \) or \( x = 14 \)
Hence, the numbers are 13 and 14.

Question. Find two consecutive positive integers, sum of whose squares is 365.
Answer: Let the two consecutive integers be \( x \) and \( x + 1 \)
ATQ \( x^2 + (x + 1)^2 = 365 \)
\( \implies \) \( x^2 + x^2 + 2x + 1 = 365 \)
\( \implies \) \( 2x^2 + 2x - 364 = 0 \)
\( \implies \) \( x^2 + x - 182 = 0 \)
\( \implies \) \( x^2 + 14x - 13x - 182 = 0 \)
\( \implies \) \( x(x + 14) - 13(x + 14) = 0 \)
\( \implies \) \( (x - 13)(x + 14) = 0 \)
\( \implies \) \( x = 13, - 14 \) (-14 is rejected because it is a negative integer)
Hence, the two consecutive positive integers are 13 and \( 13 + 1 = 14 \).

Question. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.
Answer: Let base be \( x \) cm, then height be \( (x - 7) \) cm
By Pythagoras Theorem,
\( (\text{base})^2 + (\text{height})^2 = (\text{hypotenuse})^2 \)
\( \implies \) \( x^2 + (x - 7)^2 = (13)^2 \)
\( \implies \) \( x^2 + x^2 + 49 - 14x = 169 \)
\( \implies \) \( 2x^2 - 14x - 120 = 0 \)
\( \implies \) \( x^2 - 7x - 60 = 0 \)
\( \implies \) \( x^2 - 12x + 5x - 60 = 0 \)
\( \implies \) \( x(x - 12) + 5(x - 12) = 0 \)
\( \implies \) \( (x + 5)(x - 12) = 0 \)
\( \implies \) \( x - 12 = 0 \) or \( x + 5 = 0 \)
\( \implies \) \( x = 12 \) or \( x = -5 \)
(-5 is rejected as sides can never be negative)
\( \implies \) Base = 12 cm and altitude = \( 12 - 7 = 5 \) cm

Question. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article.
Answer: Let total number of pottery articles produced in a day be \( x \)
Cost of production = Rs. \( \frac{90}{x} \)
ATQ \( 2x + 3 = \frac{90}{x} \)
\( \implies \) \( x(2x + 3) = 90 \)
\( \implies \) \( 2x^2 + 3x - 90 = 0 \)
\( \implies \) \( 2x^2 + 15x - 12x - 90 = 0 \)
\( \implies \) \( (2x + 15)(x - 6) = 0 \)
\( \implies \) \( 2x = - 15 \) or \( x - 6 = 0 \)
\( \implies \) \( x = -\frac{15}{2} \) (\( -\frac{15}{2} \) is rejected) or \( x = 6 \)
\( \therefore \) Number of articles produced per day = 6
Cost of production per article = \( \frac{90}{6} = \text{Rs. } 15 \)

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them:

Question. \( 2x^2 - 3x + 5 = 0 \)
Answer: \( 2x^2 - 3x + 5 = 0 \)
This is of the form \( ax^2 + bx + c = 0 \), where \( a = 2, b = -3 \) and \( c = 5 \)
Discriminant \( (D) = b^2 - 4ac = (-3)^2 - 4 \times 2 \times 5 = 9 - 40 = - 31 < 0 \)
Hence, no real roots exist.

Question. \( 3x^2 - 4\sqrt{3}x + 4 = 0 \)
Answer: \( 3x^2 - 4\sqrt{3}x + 4 = 0 \)
This is of the form \( ax^2 + bx + c = 0 \), where \( a = 3, b = - 4\sqrt{3} \) and \( c = 4 \)
Discriminant \( (D) = b^2 - 4ac = (-4\sqrt{3})^2 - 4 \times 3 \times 4 = 48 - 48 = 0 \)
Since, \( D = 0 \)
Hence, real and equal roots exist.
\( x = \frac{-b \pm \sqrt{D}}{2a} \)
\( \implies \) \( x = \frac{-(-4\sqrt{3}) \pm 0}{2 \times 3} = \frac{4\sqrt{3}}{6} \)
\( \therefore \) Roots are \( \frac{2\sqrt{3}}{3}, \frac{2\sqrt{3}}{3} = \frac{2}{\sqrt{3}}, \frac{2}{\sqrt{3}} \).

Question. \( 2x^2 - 6x + 3 = 0 \)
Answer: \( 2x^2 - 6x + 3 = 0 \)
This is of the form \( ax^2 + bx + c = 0 \), where \( a = 2, b = -6 \) and \( c = 3 \)
Discriminant \( (D) = b^2 - 4ac = (-6)^2 - 4 \times 2 \times 3 = 36 - 24 = 12 > 0 \)
Hence, distinct real roots exist.
\( x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-(-6) \pm \sqrt{12}}{2 \times 2} \)
\( \implies \) \( \frac{6 \pm \sqrt{4 \times 3}}{4} = \frac{6 \pm 2\sqrt{3}}{4} \)
\( = \frac{2(3 \pm \sqrt{3})}{4} \) or \( \frac{3 \pm \sqrt{3}}{2} \)
Hence, roots are \( \frac{3 + \sqrt{3}}{2}, \frac{3 - \sqrt{3}}{2} \).

Find the values of k for each of the following quadratic equations, so that they have two equal roots. 

Question. \( 2x^2 + kx + 3 = 0 \)
Answer: \( 2x^2 + kx + 3 = 0 \)
This is of the form \( ax^2 + bx + c = 0 \), where, \( a = 2, b = k \) and \( c = 3 \)
Discriminant \( (D) = b^2 - 4ac = k^2 - 4 \times 2 \times 3 = k^2 - 24 \)
For equal roots, \( D = 0 \)
\( \implies \) \( k^2 - 24 = 0 \)
\( \implies \) \( k^2 = 24 \) or \( k = \pm \sqrt{24} \)
\( \implies \) \( k = \pm \sqrt{4 \times 6} = \pm 2\sqrt{6} \)

Question. \( kx (x - 2) + 6 = 0 \)
Answer: \( kx^2 - 2kx + 6 = 0 \)
This is of the form \( ax^2 + bx + c = 0 \), where \( a = k, b = -2k \) and \( c = 6 \)
Discriminant \( (D) = b^2 - 4ac = (-2k)^2 - 4 \times k \times 6 = 4k^2 - 24k \)
For equal roots, \( D = 0 \)
\( \implies \) \( 4k^2 - 24k = 0 \)
\( \implies \) \( k(4k - 24) = 0 \)
\( \implies \) \( k = 0 \) (not possible) or \( 4k - 24 = 0 \)
\( \implies \) \( 4k = 24 \) \( \implies \) \( k = \frac{24}{4} = 6 \)

Question. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m\( ^2 \)? If so, find its length and breadth.
Answer: Let breadth of the rectangular mango grove be \( x \) m
then, the length of rectangular mango grove be \( 2x \) m
ATQ \( x \times 2x = 800 \) or \( 2x^2 = 800 \)
\( \implies \) \( x^2 = 400 \)
\( \implies \) \( x = \pm 20 \) [-20 is rejected]
Hence, breadth = 20 m and length = \( 2 \times 20 = 40 \) m
So, it is possible to design a rectangular mango grove whose length is twice its breadth.

Question. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Answer: Let the present age of one friend be \( x \) years
then the present age of other friend be \( (20 - x) \) years
4 years ago one friend’s age was \( (x - 4) \) years
4 years ago other friend’s age was \( (20 - x - 4) = (16 - x) \) years
ATQ \( (x - 4)(16 - x) = 48 \)
\( \implies \) \( 16x - x^2 - 64 + 4x = 48 \)
\( \implies \) \( x^2 - 20x + 112 = 0 \)
This is of the form \( ax^2 + bx + c = 0 \), where, \( a = 1, b = -20 \) and \( c = 112 \)
Discriminant \( (D) = b^2 - 4ac = (-20)^2 - 4 \times 1 \times 112 = 400 - 448 = - 48 < 0 \)
\( \because \) No real roots exist
\( \therefore \) The given situation is not possible.

Question. Is it possible to design a rectangular park of perimeter 80 m and area 400 m\( ^2 \)? If so, find its length and breadth.
Answer: Let the length of rectangular park be \( x \) m and breadth be \( y \) m.
Given, area = 400 m\( ^2 \)
\( \implies \) \( xy = 400 \) \( \implies \) \( y = \frac{400}{x} \)
Perimeter = 80 m
\( \implies \) \( 2(x + y) = 80 \)
\( \implies \) \( x + y = 40 \) ...(i)
Putting the value of \( y \) in equation (i), we get
\( x + \frac{400}{x} = 40 \)
\( \implies \) \( x^2 + 400 = 40x \)
\( \implies \) \( x^2 - 40x + 400 = 0 \)
\( \implies \) \( x^2 - 20x - 20x + 400 = 0 \)
\( \implies \) \( x(x - 20) - 20(x - 20) = 0 \)
\( \implies \) \( (x - 20)(x - 20) = 0 \)
\( \implies \) \( x - 20 = 0 \) or \( x = 20 \)
Since \( y = \frac{400}{x} \)
\( \implies \) \( y = \frac{400}{20} = 20 \)
\( \therefore \) Length of park = 20 m and breadth of park = 20 m
So, the given situation is possible.

Question. Which of the following is not a quadratic equation?
(a) \( 2(x - 1)^2 = 4x^2 - 2x + 1 \)
(b) \( 2x - x^2 = x^2 + 5 \)
(c) \( (\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x \)
(d) \( (x^2 + 2x)^2 = x^4 + 3 + 4x^3 \)
Answer: (c) \( (\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x \)
\( 2x^2 + 3 + 2\sqrt{6}x + x^2 = 3x^2 - 5x \)
\( \implies \) \( 2\sqrt{6}x + 5x + 3 = 0 \)

Question. Which of the following equations has two distinct real roots ?
(a) \( 2x^2 - 3\sqrt{2}x + \frac{9}{4} = 0 \)
(b) \( x^2 + x - 5 = 0 \)
(c) \( x^2 + 3x + 2\sqrt{2} = 0 \)
(d) \( 5x^2 - 3x + 1 = 0 \)
Answer: (b) \( x^2 + x - 5 = 0 \)
\( D > 0 \)

Question. Which of the following equations has no real roots ?
(a) \( x^2 - 4x + 3\sqrt{2} = 0 \)
(b) \( x^2 + 4x - 3\sqrt{2} = 0 \)
(c) \( x^2 - 4x - 3\sqrt{2} = 0 \)
(d) \( 3x^2 + 4\sqrt{3}x + 4 = 0 \)
Answer: (a) \( x^2 - 4x + 3\sqrt{2} = 0 \)
\( D < 0 \)

Question. \( (x^2 + 1)^2 - x^2 = 0 \) has
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real root
Answer: (c) no real roots

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.
Answer: Let present age of Asha be \( x \) years and present age of Nisha be \( y \) years
ATQ \( x = y^2 + 2 \)
Difference in ages = \( (x - y) \) years
Mother’s age after \( (x - y) \) years is \( x + (x - y) = 10y - 1 \)
\( \implies \) \( 2x - y - 10y + 1 = 0 \)
\( \implies \) \( 2(y^2 + 2) - 11y + 1 = 0 \)
\( \implies \) \( 2y^2 + 4 - 11y + 1 = 0 \)
\( \implies \) \( 2y^2 - 11y + 5 = 0 \)
\( \implies \) \( 2y^2 - 10y - y + 5 = 0 \)
\( \implies \) \( (y - 5)(2y - 1) = 0 \)
\( \implies \) \( y = 5 \) or \( y = \frac{1}{2} \) (rejecting)
Nisha’s present age = 5 years
Asha’s present age = \( 5^2 + 2 = 27 \) years.

Question. At t minutes past 2 p.m. the time needed by the minutes hand of a clock to show 3 p.m. was found to be 3 minutes less than \( \frac{t^2}{4} \) minutes. Find t.
Answer: ATQ \( (60 - t) = \frac{t^2}{4} - 3 \)
\( \implies \) \( 240 - 4t = t^2 - 12 \)
\( \implies \) \( t^2 + 4t - 252 = 0 \)
\( \implies \) \( t^2 + 18t - 14t - 252 = 0 \)
\( \implies \) \( (t + 18)(t - 14) = 0 \)
\( \implies \) \( t = 14, - 18 \) [rejected] \( \implies \) \( t = 14 \) minutes.