CBSE Class 10 Mathematics Quadratic Equations Assignment Set 25

Very Short and Short Answer Type Questions

Question 1. Find the value of k for which the roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other.
Answer:
Let the roots be \( \alpha \) and \( \frac{1}{\alpha} \). Their product is \( \alpha \times \frac{1}{\alpha} = 1 \).
Product of roots \( = \frac{c}{a} = \frac{k}{3} \), so \( \frac{k}{3} = 1 \Rightarrow k = 3 \).
Check: \( 3x^2 - 10x + 3 = (3x - 1)(x - 3) \) has roots \( \frac{1}{3} \) and 3, which are reciprocals.
Hence \( k = 3 \).
In simple words: Roots that are reciprocals of each other have product 1.

Teacher's Note:
1. Use product of roots = c/a.
2. Check that the discriminant is positive so that the roots are real.

Question 2. If \( x = -\frac{1}{2} \) is a solution of the quadratic equation \( 3x^2 + 2kx - 3 = 0 \), find the value of k.
Answer:
Since \( x = -\frac{1}{2} \) satisfies the equation, \( 3\left(-\frac{1}{2}\right)^2 + 2k\left(-\frac{1}{2}\right) - 3 = 0 \).
\( \frac{3}{4} - k - 3 = 0 \Rightarrow k = \frac{3}{4} - 3 = -\frac{9}{4} \).
Hence \( k = -\frac{9}{4} \).
In simple words: A root of an equation makes the equation true when it is substituted for x.

Teacher's Note:
1. Substitute the given root and solve the resulting linear equation in k.
2. Take care with the sign of the squared term: \( (-\frac{1}{2})^2 = \frac{1}{4} \).

Question 3. Write the discriminant of the quadratic equation \( (x + 5)^2 = 2(5x - 3) \).
Answer:
Expand: \( x^2 + 10x + 25 = 10x - 6 \Rightarrow x^2 + 31 = 0 \), so \( a = 1,\ b = 0,\ c = 31 \).
Discriminant \( D = b^2 - 4ac = 0 - 4 \times 1 \times 31 = \) \( -124 \).
Since \( D < 0 \), the equation has no real roots.
In simple words: First bring the equation to the standard form, then use b squared minus 4ac.

Teacher's Note:
1. Write the equation in the form \( ax^2 + bx + c = 0 \) before reading a, b and c.
2. A missing x term means b = 0.

Question 4. Find the roots of the quadratic equation \( 3x^2 - 7x - 6 = 0 \).
Answer:
Split the middle term: \( 3x^2 - 9x + 2x - 6 = 0 \) (product \( 3 \times (-6) = -18 \), sum \( -7 \)).
\( 3x(x - 3) + 2(x - 3) = 0 \Rightarrow (3x + 2)(x - 3) = 0 \).
Hence \( x = 3 \) or \( x = -\frac{2}{3} \).
In simple words: Split the middle term into two numbers with product ac and sum b, then factorise.

Teacher's Note:
1. Check by substituting x = 3: 27 - 21 - 6 = 0.
2. Each factor is put equal to zero to get one root.

Question 5. In a flight of 600 km, an aircraft was slowed due to bad weather. Its average speed for the trip was reduced by 200 km/h and the time of flight increased by 30 minutes. Find the original duration of the flight.
Answer:
Let the original speed be v km/h. Original time \( = \frac{600}{v} \) hours; new speed \( v - 200 \) km/h, new time \( = \frac{600}{v - 200} \) hours.
The time increased by 30 minutes, that is \( \frac{1}{2} \) hour: \( \frac{600}{v - 200} - \frac{600}{v} = \frac{1}{2} \).
\( \frac{600 \times 200}{v(v - 200)} = \frac{1}{2} \Rightarrow v^2 - 200v - 240000 = 0 \Rightarrow (v - 600)(v + 400) = 0 \).
Speed cannot be negative, so \( v = 600 \) km/h.
Original duration \( = \frac{600}{600} = \) 1 hour.
Check: reduced speed 400 km/h gives \( \frac{600}{400} = 1.5 \) hours, which is 30 minutes more.
In simple words: Write the two times in terms of the original speed, subtract, and set the difference equal to half an hour.

Teacher's Note:
1. Convert minutes to hours before forming the equation.
2. Reject the negative root as a speed cannot be negative.

Question 6. Solve for x: \( 3x^2 + 5\sqrt{5}\,x - 10 = 0 \).
Answer:
\( 3x^2 + 5\sqrt{5}\,x - 10 = 0 \): here \( a = 3,\ b = 5\sqrt{5},\ c = -10 \). \( D = 125 + 120 = 245 = (7\sqrt{5})^2 \).
\( x = \frac{-5\sqrt{5} \pm 7\sqrt{5}}{6} \).
\( x = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3} \) or \( x = \frac{-12\sqrt{5}}{6} = -2\sqrt{5} \).
Hence \( x = \frac{\sqrt{5}}{3} \) or \( x = -2\sqrt{5} \).
In simple words: Use the quadratic formula; the discriminant 245 is a perfect square times 5, so the roots are simple.

Teacher's Note:
1. Write \( 245 = 49 \times 5 \) and take the square root as \( 7\sqrt{5} \).
2. Simplify each root separately.

Question 7. Solve for x: \( \frac{x + 1}{x - 1} + \frac{x - 2}{x + 2} = 4 - \frac{2x + 3}{x - 2} \), where \( x \neq 1, -2, 2 \).
Answer:
Left side: \( \frac{(x + 1)(x + 2) + (x - 2)(x - 1)}{(x - 1)(x + 2)} = \frac{2x^2 + 4}{(x - 1)(x + 2)} \).
Right side: \( \frac{4(x - 2) - (2x + 3)}{x - 2} = \frac{2x - 11}{x - 2} \).
Cross-multiplying: \( (2x^2 + 4)(x - 2) = (2x - 11)(x - 1)(x + 2) \).
\( 2x^3 - 4x^2 + 4x - 8 = 2x^3 - 9x^2 - 15x + 22 \Rightarrow 5x^2 + 19x - 30 = 0 \).
\( 5x^2 + 25x - 6x - 30 = 0 \Rightarrow (x + 5)(5x - 6) = 0 \).
Both values are allowed. Hence \( x = -5 \) or \( x = \frac{6}{5} \).
In simple words: Combine the fractions on each side, cross-multiply; the cubic terms cancel and a quadratic remains.

Teacher's Note:
1. Collect each side into a single fraction before cross-multiplying.
2. Check that no root makes a denominator zero.

Question 8. Write all the values of p for which the quadratic equation \( x^2 + px + 16 = 0 \) has equal roots. Find the roots of the equation so obtained.
Answer:
Equal roots need \( D = 0 \): \( p^2 - 4 \times 1 \times 16 = 0 \Rightarrow p^2 = 64 \Rightarrow p = \pm 8 \).
For \( p = 8 \): \( x^2 + 8x + 16 = (x + 4)^2 = 0 \Rightarrow x = -4, -4 \).
For \( p = -8 \): \( x^2 - 8x + 16 = (x - 4)^2 = 0 \Rightarrow x = 4, 4 \).
Hence \( p = 8 \) (roots \( -4, -4 \)) or \( p = -8 \) (roots \( 4, 4 \)).
In simple words: Equal roots mean the discriminant is zero.

Teacher's Note:
1. Both signs of p must be written.
2. The repeated root is -p/2.

Question 9. For what value of k does the quadratic equation \( kx^2 - 6x - 1 = 0 \) have no real roots?
Answer:
For no real roots, \( D < 0 \): \( (-6)^2 - 4 \times k \times (-1) < 0 \Rightarrow 36 + 4k < 0 \).
\( 4k < -36 \Rightarrow k < -9 \).
Hence the equation has no real roots when \( k < -9 \).
In simple words: No real roots means the discriminant is negative.

Teacher's Note:
1. Do not forget that c = -1 makes the term -4ac equal to +4k.
2. The answer is a range of values, not one number.

Long Answer Type Questions

Question 10. At t minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than \( \frac{t^2}{4} \) minutes. Find t.
Answer:
At t minutes past 2 pm, the minute hand still needs \( (60 - t) \) minutes to show 3 pm.
Given: \( 60 - t = \frac{t^2}{4} - 3 \Rightarrow 240 - 4t = t^2 - 12 \Rightarrow t^2 + 4t - 252 = 0 \).
\( t^2 + 18t - 14t - 252 = 0 \Rightarrow (t + 18)(t - 14) = 0 \Rightarrow t = -18 \) or \( t = 14 \).
Time cannot be negative, so \( t = 14 \).
Check: \( 60 - 14 = 46 \) and \( \frac{14^2}{4} - 3 = 49 - 3 = 46 \).
Hence \( t = 14 \).
In simple words: The time left to 3 pm is 60 minus t, and this is set equal to t squared over 4 minus 3.

Teacher's Note:
1. Reject t = -18 because the time past 2 pm cannot be negative.
2. Always check the answer in the original statement.

Question 11. In the centre of a rectangular lawn of dimensions 50 m \( \times \) 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m\( ^2 \) (see the figure, the grass border has the same width on every side). Find the length and breadth of the pond.
Answer:

Let the width of the grass border be x m. Then the pond measures \( (50 - 2x) \) m by \( (40 - 2x) \) m.
Area of lawn \( = 50 \times 40 = 2000 \) m\( ^2 \). Area of pond \( = 2000 - 1184 = 816 \) m\( ^2 \).
\( (50 - 2x)(40 - 2x) = 816 \Rightarrow 4x^2 - 180x + 2000 = 816 \Rightarrow x^2 - 45x + 296 = 0 \).
\( (x - 8)(x - 37) = 0 \Rightarrow x = 8 \) or \( x = 37 \). A border of 37 m is impossible because the lawn is only 40 m wide, so \( x = 8 \).
Pond: length \( 50 - 16 = 34 \) m and breadth \( 40 - 16 = 24 \) m. Check: \( 34 \times 24 = 816 \).
Hence the pond is 34 m long and 24 m broad.
In simple words: Call the border width x, express the pond's sides in x, and use area of lawn minus area of grass equals area of pond.

Teacher's Note:
1. The pond is in the centre, so the border is the same on opposite sides and each side loses 2x.
2. Reject x = 37 because 2x would exceed the lawn's width.

Question 12. Solve for x: \( \left(\frac{2x}{x - 5}\right)^2 + 5\left(\frac{2x}{x - 5}\right) - 24 = 0 \), \( x \neq 5 \).
Answer:
Let \( y = \frac{2x}{x - 5} \). The equation becomes \( y^2 + 5y - 24 = 0 \Rightarrow (y + 8)(y - 3) = 0 \Rightarrow y = -8 \) or \( y = 3 \).
If \( y = 3 \): \( 2x = 3(x - 5) \Rightarrow 2x = 3x - 15 \Rightarrow x = 15 \).
If \( y = -8 \): \( 2x = -8(x - 5) \Rightarrow 10x = 40 \Rightarrow x = 4 \).
Check: for \( x = 15 \), \( y = \frac{30}{10} = 3 \); for \( x = 4 \), \( y = \frac{8}{-1} = -8 \). Both satisfy the equation.
Hence \( x = 15 \) or \( x = 4 \).
In simple words: Replace the repeated fraction by a single letter, solve the simple quadratic, then go back to x.

Teacher's Note:
1. Substitution turns the equation into an ordinary quadratic in y.
2. Each value of y gives one linear equation in x.

Word Problems

Question 13. A train travels at a certain average speed for a distance of 63 km and then travels a distance of 72 km at an average speed of 6 km/h more than its original speed. If it takes 3 hours to complete the total journey, what is its original average speed?
Answer:
Let the original speed be x km/h. Time for 63 km \( = \frac{63}{x} \) h; time for 72 km at \( (x + 6) \) km/h \( = \frac{72}{x + 6} \) h.
\( \frac{63}{x} + \frac{72}{x + 6} = 3 \Rightarrow \frac{21}{x} + \frac{24}{x + 6} = 1 \Rightarrow 21(x + 6) + 24x = x(x + 6) \).
\( 45x + 126 = x^2 + 6x \Rightarrow x^2 - 39x - 126 = 0 \Rightarrow (x - 42)(x + 3) = 0 \).
A speed cannot be negative, so \( x = 42 \).
Check: \( \frac{63}{42} + \frac{72}{48} = 1.5 + 1.5 = 3 \) hours.
Hence the original average speed is 42 km/h.
In simple words: Add the two travelling times and make the total equal to 3 hours.

Teacher's Note:
1. Time = distance / speed for each part of the journey.
2. Reject the negative root.

Question 14. Determine two consecutive negative even integers whose product is 24.
Answer:
Let the two consecutive even integers be x and \( x + 2 \). Then \( x(x + 2) = 24 \Rightarrow x^2 + 2x - 24 = 0 \).
\( (x + 6)(x - 4) = 0 \Rightarrow x = -6 \) or \( x = 4 \).
The integers are negative, so \( x = -6 \) and \( x + 2 = -4 \).
Check: \( (-6) \times (-4) = 24 \).
Hence the integers are \( -6 \) and \( -4 \).
In simple words: Consecutive even integers differ by 2; the condition that both are negative picks one root.

Teacher's Note:
1. The root x = 4 would give the positive pair 4 and 6.
2. Product of two negative numbers is positive, so the pair is consistent.

Question 15. A rectangle has a length that is 2 less than 3 times its width. If the area of the rectangle is 16 square units, find its dimensions and its perimeter.
Answer:
Let the width be y units. Then the length is \( (3y - 2) \) units.
Area: \( (3y - 2)y = 16 \Rightarrow 3y^2 - 2y - 16 = 0 \Rightarrow 3y^2 - 8y + 6y - 16 = 0 \Rightarrow (3y - 8)(y + 2) = 0 \).
Width cannot be negative, so \( y = \frac{8}{3} \) and the length is \( 3 \times \frac{8}{3} - 2 = 6 \).
Check: \( 6 \times \frac{8}{3} = 16 \).
Perimeter \( = 2\left(6 + \frac{8}{3}\right) = 2 \times \frac{26}{3} = \frac{52}{3} \).
Hence the length is 6 units, the width is \( \frac{8}{3} \) units and the perimeter is \( \frac{52}{3} \) units.
In simple words: Write the length in terms of the width, use length times width equals 16, and reject the negative width.

Teacher's Note:
1. Add length and width first, then double it for the perimeter.
2. The width 8/3 is not a whole number, which is fine because the sides are lengths.

Question 16. The difference of two numbers is 2 and their product is 224. Find the numbers.
Answer:
Let the smaller number be x; the other is \( x + 2 \). Then \( x(x + 2) = 224 \Rightarrow x^2 + 2x - 224 = 0 \).
\( x^2 + 16x - 14x - 224 = 0 \Rightarrow (x + 16)(x - 14) = 0 \Rightarrow x = -16 \) or \( x = 14 \).
If \( x = 14 \), the numbers are 14 and 16. If \( x = -16 \), the numbers are -16 and -14.
Check: \( 14 \times 16 = 224 \) and \( (-16) \times (-14) = 224 \).
Hence the numbers are 14 and 16 (or -16 and -14 if negative numbers are allowed).
In simple words: Two numbers with difference 2 can be written as x and x plus 2.

Teacher's Note:
1. The question does not say the numbers are positive, so both pairs are valid.
2. Factorise 224 = 14 x 16 to see the answer quickly.

Question 17. X and Y working together can complete a certain project in 6 days. If X alone works on the project, he will take 5 days less than Y to complete it. Find the number of days required for X and for Y to complete the project alone.
Answer:
Let Y alone take y days; then X alone takes \( (y - 5) \) days.
In one day they do \( \frac{1}{y - 5} + \frac{1}{y} = \frac{1}{6} \Rightarrow 6(2y - 5) = y(y - 5) \Rightarrow y^2 - 17y + 30 = 0 \).
\( (y - 15)(y - 2) = 0 \Rightarrow y = 15 \) or \( y = 2 \). The value \( y = 2 \) would make X's time \( -3 \) days, which is impossible, so \( y = 15 \).
Check: \( \frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6} \).
Hence Y alone takes 15 days and X alone takes 10 days.
In simple words: Add their one-day work and make it equal to one-sixth of the project.

Teacher's Note:
1. Work done in one day = 1 / (days taken alone).
2. Reject y = 2 since X would then need a negative number of days.

Question 18. The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, the sum of the new fraction and the original fraction is \( \frac{29}{20} \). Find the original fraction.
Answer:
Let the denominator be x. The original fraction is \( \frac{x - 3}{x} \) and the new fraction is \( \frac{x - 1}{x + 2} \).
\( \frac{x - 3}{x} + \frac{x - 1}{x + 2} = \frac{29}{20} \Rightarrow \frac{2x^2 - 2x - 6}{x(x + 2)} = \frac{29}{20} \).
\( 20(2x^2 - 2x - 6) = 29x(x + 2) \Rightarrow 11x^2 - 98x - 120 = 0 \Rightarrow (11x + 12)(x - 10) = 0 \).
\( x = 10 \) or \( x = -\frac{12}{11} \). A denominator that is a whole number is required, so \( x = 10 \).
Check: \( \frac{7}{10} + \frac{9}{12} = 0.7 + 0.75 = 1.45 = \frac{29}{20} \).
Hence the original fraction is \( \frac{7}{10} \).
In simple words: Express both fractions with the denominator x, add them, and equate to 29/20.

Teacher's Note:
1. Numerator = denominator - 3 gives the fraction (x - 3)/x.
2. Reject the fractional value of x since the denominator of a fraction of this kind is a positive integer.

Question 19. Five times a positive integer is less than twice its square by 3. Find the integer.
Answer:
Let the integer be x. Then \( 5x = 2x^2 - 3 \Rightarrow 2x^2 - 5x - 3 = 0 \).
\( 2x^2 - 6x + x - 3 = 0 \Rightarrow (2x + 1)(x - 3) = 0 \Rightarrow x = -\frac{1}{2} \) or \( x = 3 \).
The integer is positive, so \( x = 3 \). Check: \( 5 \times 3 = 15 \) and \( 2 \times 9 - 3 = 15 \).
Hence the integer is 3.
In simple words: Translate 'less than twice its square by 3' as 5x = 2x squared minus 3.

Teacher's Note:
1. The root -1/2 is neither positive nor an integer.
2. Check the answer in the original sentence.

Question 20. If Zeba were younger by 5 years than what she really is, the square of her age would have been 11 more than five times her actual age. What is her age now?
Answer:
Let her present age be x years. Her age 5 years ago would be \( (x - 5) \) years.
\( (x - 5)^2 = 5x + 11 \Rightarrow x^2 - 10x + 25 = 5x + 11 \Rightarrow x^2 - 15x + 14 = 0 \).
\( (x - 1)(x - 14) = 0 \Rightarrow x = 1 \) or \( x = 14 \). If \( x = 1 \), her age 5 years ago would be \( -4 \), which is impossible, so \( x = 14 \).
Check: \( (14 - 5)^2 = 81 \) and \( 5 \times 14 + 11 = 81 \).
Hence her present age is 14 years.
In simple words: The younger age is x minus 5; square it and compare with five times the actual age plus 11.

Teacher's Note:
1. An age 5 years ago must be positive, which removes x = 1.
2. Check by squaring 9 and comparing with 81.