GSEB Class 10 Maths Solutions Chapter 4 દ્વિઘાત સમીકરણ Exercise 4.3

Question 1. Find the roots of the following quadratic equations, if they exist by the method of completing the square:
(i) \( 2x^2 - 7x + 3 = 0 \)
(ii) \( 2x^2 + x - 4 = 0 \)
(iii) \( 4x^2 + 4\sqrt{3}x + 3 = 0 \)
(iv) \( 2x^2 + x + 4 = 0 \)
Answer:
(i) The roots are \( 3 \) and \( \frac{1}{2} \).
(ii) The roots are \( \frac{-1+\sqrt{33}}{4} \) and \( \frac{-1-\sqrt{33}}{4} \).
(iii) The roots are \( \frac{-\sqrt{3}}{2} \) and \( \frac{-\sqrt{3}}{2} \).
(iv) The given quadratic equation has no real roots.
In simple words: To solve by completing the square, rearrange the equation to isolate the variable terms, add the square of half the coefficient of x to both sides, and solve for x.

Exam Tip: Always check the discriminant (\( b^2 - 4ac \)) first; if it is negative, the equation has no real roots and you can stop immediately.

Question 2. Find the roots of the quadratic equations given in Q.1, above by applying the quadratic formula.
Answer:
(i) For \( 2x^2 - 7x + 3 = 0 \), the roots are \( 3 \) and \( \frac{1}{2} \).
(ii) For \( 2x^2 + x - 4 = 0 \), the roots are \( \frac{-1+\sqrt{33}}{4} \) and \( \frac{-1-\sqrt{33}}{4} \).
(iii) For \( 4x^2 + 4\sqrt{3}x + 3 = 0 \), the roots are \( \frac{-\sqrt{3}}{2} \) and \( \frac{-\sqrt{3}}{2} \).
(iv) For \( 2x^2 + x + 4 = 0 \), there are no real roots.
In simple words: Use the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the roots directly using the coefficients a, b, and c from the equation.

Exam Tip: Ensure you correctly identify the signs of a, b, and c before plugging them into the quadratic formula.

Question 3. Find the roots of the following equations:
(i) \( x - \frac{1}{x} = 3, x \neq 0 \)
(ii) \( \frac{1}{x+4} - \frac{1}{x-7} = \frac{11}{30}, x \neq -4, 7 \)
Answer:
(i) The roots are \( \frac{3+\sqrt{13}}{2} \) and \( \frac{3-\sqrt{13}}{2} \).
(ii) The roots are \( 2 \) and \( 1 \).
In simple words: First, simplify the equations into the standard quadratic form \( ax^2 + bx + c = 0 \), then solve using the quadratic formula or factorization.

Exam Tip: When dealing with fractions, multiply the entire equation by the common denominator to clear the fractions before solving.

Question 4. The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is \( \frac{1}{3} \). Find his present age.
Answer: Let the present age of Rehman be \( x \) years. According to the problem, \( \frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3} \). Solving this leads to the quadratic equation \( x^2 - 4x - 21 = 0 \). Factoring gives \( (x-7)(x+3) = 0 \). Since age cannot be negative, \( x = 7 \). Hence, the present age of Rehman is 7 years.
In simple words: Set up an equation based on his age 3 years ago and 5 years from now, then solve the resulting quadratic equation for x.

Exam Tip: Always discard negative values for age as they are physically impossible in this context.

Question 5. In a class, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product would have been 210. Find her marks in the two subjects.
Answer: Let marks in Mathematics be \( x \), then marks in English are \( 30-x \). The condition \( (x+2)(30-x-3) = 210 \) simplifies to \( x^2 - 25x + 156 = 0 \). Solving this gives \( x = 13 \) or \( x = 12 \). If marks in Mathematics are 13, English marks are 17. If marks in Mathematics are 12, English marks are 18.
In simple words: Create an equation using the given conditions for the marks, solve the quadratic equation, and find the two possible sets of marks.

Exam Tip: Since there are two possible values for x, provide both valid pairs of marks as the final answer.

Question 6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Answer: Let the shorter side be \( x \) m. Then the longer side is \( x+30 \) m and the diagonal is \( x+60 \) m. By Pythagoras theorem, \( x^2 + (x+30)^2 = (x+60)^2 \). This simplifies to \( x^2 - 60x - 2700 = 0 \). Solving gives \( x = 90 \) (since \( x = -30 \) is rejected). Thus, the shorter side is 90 m and the longer side is 120 m.
In simple words: Use the Pythagorean theorem for the rectangle's sides and diagonal to form a quadratic equation, then solve for the shorter side.

Exam Tip: Remember that in a rectangle, the diagonal forms a right-angled triangle with the two sides.

Question 7. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, would have taken 1 hour less for the same journey. Find the speed of the train.
Answer: Let the speed be \( x \) km/h. The time taken is \( \frac{360}{x} \). If speed is \( x+5 \), time is \( \frac{360}{x+5} \). Given \( \frac{360}{x} - \frac{360}{x+5} = 1 \), this leads to \( x^2 + 5x - 1800 = 0 \). Solving gives \( x = 40 \) (rejecting negative speed). The speed of the train is 40 km/h.
In simple words: Use the relationship \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \) to set up the equation and solve for the speed.

Exam Tip: Always check that your final speed value makes sense in the context of the problem (e.g., it must be positive).

Question 8. Two water taps together can fill a tank in \( 9 \frac{3}{8} \) hours. The larger tap takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer: Let the larger tap take \( x \) hours and the smaller take \( x+10 \) hours. Together they fill the tank in \( \frac{75}{8} \) hours. The equation is \( \frac{1}{x} + \frac{1}{x+10} = \frac{8}{75} \), which simplifies to \( 4x^2 - 35x - 375 = 0 \). Solving gives \( x = 15 \) (rejecting negative time). The larger tap takes 15 hours and the smaller takes 25 hours.
In simple words: Calculate the rate of each tap (part of the tank filled per hour) and set their sum equal to the combined rate.

Exam Tip: Ensure you convert the mixed fraction \( 9 \frac{3}{8} \) to an improper fraction \( \frac{75}{8} \) before setting up the equation.

Question 9. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Answer: Let the speed of the passenger train be \( x \) km/h and the express train be \( x+11 \) km/h. The equation is \( \frac{132}{x} - \frac{132}{x+11} = 1 \), which simplifies to \( x^2 + 11x - 1452 = 0 \). Solving gives \( x = 33 \). The passenger train speed is 33 km/h and the express train speed is 44 km/h.
In simple words: Use the time difference to create an equation based on the speeds of the two trains and solve for x.

Exam Tip: Double-check your quadratic formula calculations, as large numbers like 1452 can lead to simple arithmetic errors.

Question 10. Sum of the area of the two squares is 468 m\(^2\). If the difference of their perimeters is 24 m, find the sides of the two squares.
Answer: Let the sides be \( x \) and \( y \). We have \( x^2 + y^2 = 468 \) and \( 4x - 4y = 24 \), so \( x - y = 6 \) or \( x = y+6 \). Substituting gives \( (y+6)^2 + y^2 = 468 \), which simplifies to \( y^2 + 6y - 216 = 0 \). Solving gives \( y = 12 \). Then \( x = 18 \). The sides are 18 m and 12 m.
In simple words: Use the area and perimeter formulas to create a system of equations, then substitute one into the other to solve.

Exam Tip: Clearly define your variables at the start to avoid confusion between the two different squares.

Question 7. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, would have taken 1 hour less for the same journey. Find the speed of the train.
Answer: Let the speed of the train be \( x \) km/h. Distance = 360 km. Time taken = \( \frac{360}{x} \) h. If the speed were 5 km/h more, the time taken would be \( \frac{360}{x+5} \) h. According to the problem, \( \frac{360}{x} - \frac{360}{x+5} = 1 \). Solving this, \( 360(x+5-x) = x(x+5) \), so \( x^2 + 5x - 1800 = 0 \). Factoring gives \( (x+45)(x-40) = 0 \). Since speed cannot be negative, \( x = 40 \). The speed of the train is 40 km/h.
In simple words: We use the formula that time equals distance divided by speed. By setting up an equation for the two different speeds, we find the original speed is 40 km/h.

Exam Tip: Always remember that speed must be a positive value, so discard any negative roots obtained from the quadratic equation.

Question 8. Two water taps together can fill a tank in hours. The larger tap takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Answer: Let the time taken by the larger tap be \( x \) hours. Then the smaller tap takes \( x+10 \) hours. In 1 hour, the larger tap fills \( \frac{1}{x} \) and the smaller fills \( \frac{1}{x+10} \). Together they fill \( \frac{1}{x} + \frac{1}{x+10} = \frac{8}{75} \). This simplifies to \( 4x^2 - 35x - 375 = 0 \). Solving for \( x \), we get \( x = 15 \) or \( x = -6.25 \). Since time cannot be negative, the larger tap takes 15 hours and the smaller tap takes 25 hours.
In simple words: We calculate the portion of the tank each tap fills in one hour. By adding these portions together, we solve for the time each tap needs.

Exam Tip: When dealing with work-rate problems, always express the work done by each agent as a fraction of the total work per unit of time.

Question 9. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore. If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.
Answer: Let the speed of the passenger train be \( x \) km/h. The speed of the express train is \( x+11 \) km/h. Time taken by the passenger train is \( \frac{132}{x} \) and by the express train is \( \frac{132}{x+11} \). Given \( \frac{132}{x} - \frac{132}{x+11} = 1 \), we get \( x^2 + 11x - 1452 = 0 \). Solving this, \( x = 33 \) or \( x = -44 \). Since speed is positive, the passenger train speed is 33 km/h and the express train speed is 44 km/h.
In simple words: We set up an equation based on the time difference between the two trains. Solving the quadratic equation gives us the speed for both.

Exam Tip: Ensure you correctly identify which train is faster to set up the time difference equation as (Time of slower train) - (Time of faster train) = Difference.

Question 10. Sum of the area of the two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.
Answer: Let the sides of the squares be \( x \) and \( y \) (where \( x > y \)). Area sum: \( x^2 + y^2 = 468 \). Perimeter difference: \( 4x - 4y = 24 \), so \( x - y = 6 \) or \( x = y + 6 \). Substituting \( x \) in the area equation: \( (y+6)^2 + y^2 = 468 \). This simplifies to \( 2y^2 + 12y - 432 = 0 \), or \( y^2 + 6y - 216 = 0 \). Factoring gives \( (y+18)(y-12) = 0 \). Since side length is positive, \( y = 12 \) m and \( x = 18 \) m.
In simple words: We use the perimeter difference to relate the two sides, then substitute this into the area sum equation to find the side lengths.

Exam Tip: Always define your variables clearly at the start and state the units (meters) in your final answer.