Let us work out 1.4 | WB Class 10

Ganit Prakash - Class-X - Let us work out 1.4

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Q1

Question 1 (i)

Let us write whether Sridhara Acharyya's formula is applicable to solve the equation \(4x^2 + (2x - 1)(2x + 1) = 4x(2x - 1)\)

Let's simplify the given equation:

\(4x^2 + (2x - 1)(2x + 1) = 4x(2x - 1)\)

or, \(4x^2 + (4x^2 - 1) = 8x^2 - 4x\)

or, \(8x^2 - 1 = 8x^2 - 4x\)

or, \(4x - 1 = 0\)

Since the highest power of the variable \(x\) is 1, this is a linear equation, not a quadratic equation.

Answer: No, Sridhara Acharyya's formula is not applicable because it is not a quadratic equation.

Question 1 (ii)

Let us write what type of equations can be solved with the help of Sridhara Acharyya's formula.

Answer: Sridhara Acharyya's formula is used to solve quadratic equations in one variable (equations of the general form \(ax^2 + bx + c = 0\), where \(a \neq 0\)).

Question 1 (iii)

By applying Sridhara Acharyya's formula in the equation \(5x^2 + 2x - 7 = 0\), it is found that \(x = \frac{k \pm 12}{10}\), let us write by calculating, what will be the value of k.

Comparing the equation \(5x^2 + 2x - 7 = 0\) with \(ax^2 + bx + c = 0\), we get:

\(a = 5\), \(b = 2\), and \(c = -7\)

Using Sridhara Acharyya's formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

\(x = \frac{-2 \pm \sqrt{2^2 - 4(5)(-7)}}{2(5)}\)

or, \(x = \frac{-2 \pm \sqrt{4 + 140}}{10}\)

or, \(x = \frac{-2 \pm \sqrt{144}}{10}\)

or, \(x = \frac{-2 \pm 12}{10}\)

Comparing this with the given expression \(x = \frac{k \pm 12}{10}\), we find:

Answer: \(k = -2\)

📗 Let us work out 1.4

Q2 (i to ix)

Question 2 (i)

If the following quadratic equations have real roots, then let us determine them with the help of Sridhara Acharyya's formula: \(3x^2 + 11x - 4 = 0\)

Comparing with \(ax^2 + bx + c = 0\), we get \(a = 3, b = 11, c = -4\).

Discriminant (\(b^2 - 4ac\)) = \((11)^2 - 4(3)(-4) = 121 + 48 = 169 > 0\). (Real roots exist)

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

or, \(x = \frac{-11 \pm \sqrt{169}}{2(3)}\)

or, \(x = \frac{-11 \pm 13}{6}\)

Either \(x = \frac{-11 + 13}{6} = \frac{2}{6} = \frac{1}{3}\)

Or \(x = \frac{-11 - 13}{6} = \frac{-24}{6} = -4\)

Answer: The roots are \(\frac{1}{3}\) and \(-4\).

Question 2 (ii)

\( (x - 2)(x + 4) + 9 = 0 \)

Simplifying: \(x^2 + 4x - 2x - 8 + 9 = 0 \implies x^2 + 2x + 1 = 0\)

Here, \(a = 1, b = 2, c = 1\).

Discriminant (\(b^2 - 4ac\)) = \((2)^2 - 4(1)(1) = 4 - 4 = 0\). (Real and equal roots exist)

\(x = \frac{-2 \pm \sqrt{0}}{2(1)}\)

or, \(x = \frac{-2}{2} = -1\)

Answer: The roots are \(-1\) and \(-1\).

Question 2 (iii)

\( (4x - 3)^2 - 2(x + 3) = 0 \)

Simplifying: \(16x^2 - 24x + 9 - 2x - 6 = 0 \implies 16x^2 - 26x + 3 = 0\)

Here, \(a = 16, b = -26, c = 3\).

Discriminant (\(b^2 - 4ac\)) = \((-26)^2 - 4(16)(3) = 676 - 192 = 484 > 0\). (Real roots exist)

\(x = \frac{-(-26) \pm \sqrt{484}}{2(16)}\)

or, \(x = \frac{26 \pm 22}{32}\)

Either \(x = \frac{26 + 22}{32} = \frac{48}{32} = \frac{3}{2}\)

Or \(x = \frac{26 - 22}{32} = \frac{4}{32} = \frac{1}{8}\)

Answer: The roots are \(\frac{3}{2}\) and \(\frac{1}{8}\).

Question 2 (iv)

\( 3x^2 + 2x - 1 = 0 \)

Here, \(a = 3, b = 2, c = -1\).

Discriminant (\(b^2 - 4ac\)) = \((2)^2 - 4(3)(-1) = 4 + 12 = 16 > 0\). (Real roots exist)

\(x = \frac{-2 \pm \sqrt{16}}{2(3)}\)

or, \(x = \frac{-2 \pm 4}{6}\)

Either \(x = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3}\)

Or \(x = \frac{-2 - 4}{6} = \frac{-6}{6} = -1\)

Answer: The roots are \(\frac{1}{3}\) and \(-1\).

Question 2 (v)

\( 3x^2 + 2x + 1 = 0 \)

Here, \(a = 3, b = 2, c = 1\).

Discriminant (\(b^2 - 4ac\)) = \((2)^2 - 4(3)(1) = 4 - 12 = -8 < 0\).

Since the discriminant is negative, there are no real roots.

Answer: No real roots exist.

Question 2 (vi)

\( 10x^2 - x - 3 = 0 \)

Here, \(a = 10, b = -1, c = -3\).

Discriminant (\(b^2 - 4ac\)) = \((-1)^2 - 4(10)(-3) = 1 + 120 = 121 > 0\). (Real roots exist)

\(x = \frac{-(-1) \pm \sqrt{121}}{2(10)}\)

or, \(x = \frac{1 \pm 11}{20}\)

Either \(x = \frac{1 + 11}{20} = \frac{12}{20} = \frac{3}{5}\)

Or \(x = \frac{1 - 11}{20} = \frac{-10}{20} = -\frac{1}{2}\)

Answer: The roots are \(\frac{3}{5}\) and \(-\frac{1}{2}\).

Question 2 (vii)

\( 10x^2 - x + 3 = 0 \)

Here, \(a = 10, b = -1, c = 3\).

Discriminant (\(b^2 - 4ac\)) = \((-1)^2 - 4(10)(3) = 1 - 120 = -119 < 0\).

Since the discriminant is negative, there are no real roots.

Answer: No real roots exist.

Question 2 (viii)

\( 25x^2 - 30x + 7 = 0 \)

Here, \(a = 25, b = -30, c = 7\).

Discriminant (\(b^2 - 4ac\)) = \((-30)^2 - 4(25)(7) = 900 - 700 = 200 > 0\). (Real roots exist)

\(x = \frac{-(-30) \pm \sqrt{200}}{2(25)}\)

or, \(x = \frac{30 \pm 10\sqrt{2}}{50}\)

or, \(x = \frac{10(3 \pm \sqrt{2})}{50}\)

or, \(x = \frac{3 \pm \sqrt{2}}{5}\)

Answer: The roots are \(\frac{3 + \sqrt{2}}{5}\) and \(\frac{3 - \sqrt{2}}{5}\).

Question 2 (ix)

\( (4x - 2)^2 + 6x = 25 \)

Simplifying: \(16x^2 - 16x + 4 + 6x - 25 = 0 \implies 16x^2 - 10x - 21 = 0\)

Here, \(a = 16, b = -10, c = -21\).

Discriminant (\(b^2 - 4ac\)) = \((-10)^2 - 4(16)(-21) = 100 + 1344 = 1444 > 0\). (Real roots exist)

\(\sqrt{1444} = 38\)

or, \(x = \frac{-(-10) \pm 38}{2(16)}\)

or, \(x = \frac{10 \pm 38}{32}\)

Either \(x = \frac{10 + 38}{32} = \frac{48}{32} = \frac{3}{2}\)

Or \(x = \frac{10 - 38}{32} = \frac{-28}{32} = -\frac{7}{8}\)

Answer: The roots are \(\frac{3}{2}\) and \(-\frac{7}{8}\).

📗 Let us work out 1.4

Q3 (i to ix)

Question 3 (i)

Sathi has drawn a right-angled triangle whose length of the hypotenuse is 6cm. more than the twice of that of the shortest side. If the length of the third side is 2 cm. less than the length of the hypotenuse, then by calculating, let us write the lengths of three sides of the right-angled triangle drawn by Sathi.

Let the shortest side be \(x\text{ cm}\).

Then, hypotenuse = \((2x + 6)\text{ cm}\).

Third side = Hypotenuse - 2 = \((2x + 6) - 2 = (2x + 4)\text{ cm}\).

Applying Pythagoras Theorem: \((Shortest \ Side)^2 + (Third \ Side)^2 = (Hypotenuse)^2\)

\(x^2 + (2x + 4)^2 = (2x + 6)^2\)

or, \(x^2 + 4x^2 + 16x + 16 = 4x^2 + 24x + 36\)

or, \(x^2 - 8x - 20 = 0\)

Using Sridhara Acharyya's formula: \(a=1, b=-8, c=-20\)

or, \(x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-20)}}{2(1)} = \frac{8 \pm \sqrt{64 + 80}}{2} = \frac{8 \pm \sqrt{144}}{2} = \frac{8 \pm 12}{2}\)

either \(x = \frac{20}{2} = 10\) or \(x = \frac{-4}{2} = -2\).

Since length cannot be negative, \(x = 10\)

Shortest side = \(10\text{ cm}\).

Third side = \(2(10) + 4 = 24\text{ cm}\).

Hypotenuse = \(2(10) + 6 = 26\text{ cm}\).

Answer: The sides are \(10\text{ cm}\), \(24\text{ cm}\), and \(26\text{ cm}\).

Question 3 (ii)

If a two digit positive number is multiplied by its unit digit, then the product is 189 and if the tens digit is twice the unit digit, then let us calculate the unit digit.

Let the unit digit be \(x\).

Then the tens digit is \(2x\).

The two-digit number is \(10 \times (\text{tens digit}) + (\text{unit digit}) = 10(2x) + x = 20x + x = 21x\)

According to the condition:

\(\text{Number} \times \text{unit digit} = 189\)

or, \(21x \times x = 189\)

or, \(21x^2 = 189\)

or, \(x^2 = \frac{189}{21} = 9\)

\(x = \pm 3\)

Since it is a positive number, the digit \(x\) must be positive. Therefore, \(x = 3\).

Answer: The unit digit is \(3\).

Question 3 (iii)

The speed of Salma is 1 m/second more than the speed of Anik. In a 180 m race, Salma reaches 2 seconds before than Anik. Let us write by calculating the speed of Anik in m/sec.

Let Anik's speed be \(x\text{ m/sec}\).

Then Salma's speed is \((x + 1)\text{ m/sec}\).

Time taken by Anik to cover 180 m = \(\frac{180}{x}\text{ seconds}\).

Time taken by Salma to cover 180 m = \(\frac{180}{x + 1}\text{ seconds}\).

According to the problem, Salma takes 2 seconds less than Anik:

\(\frac{180}{x} - \frac{180}{x + 1} = 2\)

or, \(\frac{180(x + 1) - 180x}{x(x + 1)} = 2\)

or, \(\frac{180}{x^2 + x} = 2\)

or, \(2x^2 + 2x = 180 \implies x^2 + x - 90 = 0\)

Using factorization: \(x^2 + 10x - 9x - 90 = 0 \implies x(x + 10) - 9(x + 10) = 0 \implies (x - 9)(x + 10) = 0\)

So, \(x = 9\) or \(x = -10\).

Speed cannot be negative, so \(x = 9\).

Answer: Anik's speed is \(9\text{ m/sec}\).

Question 3 (iv)

There is a square park in our locality. The area of a rectangular park is 78 sqm. less than the twice of the area of that square-shaped park whose length is 5m. more than the length of the side of that park and the breadth is 3m. less than the length of the side of that park. Let us write by calculating, the length of the side of the square-shaped park.

Let the side length of the square park be \(x\text{ m}\).

Area of the square park = \(x^2\text{ sq.m}\).

For the rectangular park: length = \((x + 5)\text{ m}\), breadth = \((x - 3)\text{ m}\).

Area of rectangular park = \((x + 5)(x - 3)\text{ sq.m}\).

According to the condition:

\((x + 5)(x - 3) = 2x^2 - 78\)

or, \(x^2 - 3x + 5x - 15 = 2x^2 - 78\)

or, \(x^2 + 2x - 15 = 2x^2 - 78\)

or, \(x^2 - 2x - 63 = 0\)

or, \(x^2 - 9x + 7x - 63 = 0 \implies x(x - 9) + 7(x - 9) = 0 \implies (x - 9)(x + 7) = 0\)

Since side length must be positive, \(x = 9\)

Answer: The length of the side of the square park is \(9\text{ m}\).

Question 3 (v)

In our village, Proloy babu bought 350 chilli plants for planting in his rectangular land. When he put the plants in rows, he noticed that, if he would put 24 plants more than the number of rows in each row, 10 plants would remain excess. Let us write by calculating the number of rows.

Let the number of rows be \(x\).

Number of plants in each row = \(x + 24\).

Total plants planted in rows = \(x(x + 24)\)

According to the condition, after planting, 10 plants are left over from the total 350 plants:

\(x(x + 24) + 10 = 350\)

or, \(x^2 + 24x - 340 = 0\)

Using Sridhara Acharyya's formula: \(a=1, b=24, c=-340\)

or, \(x = \frac{-24 \pm \sqrt{(24)^2 - 4(1)(-340)}}{2(1)} = \frac{-24 \pm \sqrt{576 + 1360}}{2} = \frac{-24 \pm \sqrt{1936}}{2}\)

or, \(x = \frac{-24 \pm 44}{2}\)

either \(x = \frac{20}{2} = 10\) or \(x = \frac{-68}{2} = -34\).

Since the number of rows cannot be negative, \(x = 10\).

Answer: The number of rows is \(10\).

Question 3 (vi)

Joseph and Kuntal work in a factory. Joseph takes 5 minutes less time than Kuntal to make a product. Joseph makes 6 products more than Kuntal while working for 6 hours. Let us write by calculating, the number of products Kuntal makes during that time.

Let the time taken by Kuntal to make one product be \(x\text{ minutes}\).

Then, the time taken by Joseph to make one product is \((x - 5)\text{ minutes}\).

Total working time = 6 hours = \(6 \times 60 = 360\text{ minutes}\).

Number of products Kuntal makes in 360 mins = \(\frac{360}{x}\).

Number of products Joseph makes in 360 mins = \(\frac{360}{x - 5}\).

According to the condition:

\(\frac{360}{x - 5} - \frac{360}{x} = 6\)

or, \(\frac{360x - 360(x - 5)}{x(x - 5)} = 6\)

or, \(\frac{1800}{x^2 - 5x} = 6 \implies 6(x^2 - 5x) = 1800 \implies x^2 - 5x - 300 = 0\)

or, \(x^2 - 20x + 15x - 300 = 0 \implies x(x - 20) + 15(x - 20) = 0 \implies (x - 20)(x + 15) = 0\)

Since time cannot be negative, \(x = 20\).

Kuntal takes 20 minutes to make one product.

Number of products Kuntal makes in 6 hours = \(\frac{360}{20} = 18\).

Answer: Kuntal makes \(18\) products.

Question 3 (vii)

The speed of a boat in still water is 8 kms/hr. If the boat can go 15 kms. down stream and 22kms. up stream in 5 hours, then let us write by calculating, the speed of the stream.

Let the speed of the stream be \(x\text{ km/hr}\).

Speed of the boat downstream = \((8 + x)\text{ km/hr}\).

Speed of the boat upstream = \((8 - x)\text{ km/hr}\).

Time taken for 15 km downstream = \(\frac{15}{8 + x}\text{ hours}\).

Time taken for 22 km upstream = \(\frac{22}{8 - x}\text{ hours}\).

Total time = 5 hours:

\(\frac{15}{8 + x} + \frac{22}{8 - x} = 5\)

or, \(\frac{15(8 - x) + 22(8 + x)}{(8 + x)(8 - x)} = 5\)

or, \(\frac{120 - 15x + 176 + 22x}{64 - x^2} = 5\)

or, \(296 + 7x = 5(64 - x^2)\)

or, \(296 + 7x = 320 - 5x^2\)

or, \(5x^2 + 7x - 24 = 0\)

or, \(5x^2 + 15x - 8x - 24 = 0 \implies 5x(x + 3) - 8(x + 3) = 0 \implies (x + 3)(5x - 8) = 0\)

Since speed cannot be negative, \(x = \frac{8}{5} = 1.6\)

Answer: The speed of the stream is \(1.6\text{ km/hr}\).

Question 3 (viii)

A superfast train runs at the speed of 15kms/hr more than that of an express train. Leaving the same station the superfast train reached a station of 180kms. distance 1 hour before than the express train. Let us determine the speed of the superfast train in km/hr.

Let the speed of the express train be \(x\text{ km/hr}\).

Then, the speed of the superfast train is \((x + 15)\text{ km/hr}\).

Time taken by express train to cover 180 km = \(\frac{180}{x}\text{ hours}\).

Time taken by superfast train to cover 180 km = \(\frac{180}{x + 15}\text{ hours}\).

According to the condition:

\(\frac{180}{x} - \frac{180}{x + 15} = 1\)

or, \(\frac{180(x + 15) - 180x}{x(x + 15)} = 1\)

or, \(\frac{2700}{x^2 + 15x} = 1 \implies x^2 + 15x - 2700 = 0\)

or, \(x^2 + 60x - 45x - 2700 = 0 \implies x(x + 60) - 45(x + 60) = 0 \implies (x - 45)(x + 60) = 0\)

Since speed cannot be negative, \(x = 45\).

Speed of express train = \(45\text{ km/hr}\).

Speed of superfast train = \(45 + 15 = 60\text{ km/hr}\).

Answer: The speed of the superfast train is \(60\text{ km/hr}\).

Question 3 (ix)

Rehana went to the market and saw that the price of dal of 1 kg. is Rs. 20 and the price of rice of 1 kg. is Rs. 40 less than that of price of 1 kg fish. The total quantity of fish and that of dal each in Rs. 240 is equal to the quantity of rice in Rs. 280. Let us calculate the cost price of 1 kg. fish.

Let the price of 1 kg of fish be ₹ \(x\).

Price of 1 kg of rice = ₹ \((x - 40)\).

Note: There is a known misprint in the standard English textbook for this problem where the price of dal is printed as ₹20 instead of ₹60. The calculation below is shown by assuming the intended correct price of dal is ₹60 per kg to obtain a valid integer answer.

Quantity of fish bought for ₹ 240 = \(\frac{240}{x}\text{ kg}\).

Quantity of dal bought for ₹ 240 = \(\frac{240}{60} = 4\text{ kg}\).

Quantity of rice bought for ₹ 280 = \(\frac{280}{x - 40}\text{ kg}\).

According to the condition:

\(\frac{240}{x} + 4 = \frac{280}{x - 40}\)

Dividing the entire equation by 4:

or, \(\frac{60}{x} + 1 = \frac{70}{x - 40}\)

or, \(1 = \frac{70}{x - 40} - \frac{60}{x}\)

or, \(1 = \frac{70x - 60(x - 40)}{x(x - 40)}\)

or, \(x(x - 40) = 70x - 60x + 2400\)

or, \(x^2 - 40x = 10x + 2400\)

or, \(x^2 - 50x - 2400 = 0\)

or, \(x^2 - 80x + 30x - 2400 = 0 \implies x(x - 80) + 30(x - 80) = 0 \implies (x - 80)(x + 30) = 0\)

Since price cannot be negative, \(x = 80\).

Answer: The cost price of 1 kg fish is ₹ \(80\).

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WBBSE Board — Class 10 Quadratic Equations - Ganit Prakash - Let us work out 1.4 Solutions.

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