WBBSE Solutions For Class 10 Maths Chapter 1 Quadratic Equations In One Variable Exercise 1.4
Question 1. Let us write by understanding whether Sreedhar Acharya’s formula is applicable or not applicable to solve the equation 4x2 + (2x-1) (2x + 1) = 4x (2x-1).
Solution: 4x2 + (2x-1) (2x + 1) = 4x (2x-1)
=> 4x2 + 4x2 – 1 = 8x2 – 4x
=> 8x2-8x2+4x-1=0
=> 4x-1=0
It is a linear equation of one variable. So, it is not possible to apply Sreedhar Acharya’s formula.
Read and Learn More WBBSE Solutions For Class 10 Maths
Question 2. Let us write by understanding what type of equation can be solved with the help of Sreedhar Acharyya’s formula.
Solution: Quadratic equation in one variable.
Question 3. By applying Sreedhar Acharyya’s formula in equation 5×2 + 2x-7= 0, it is found that x = k±12/10; let us write by calculating what will be the value of k.
Solution: Applying Sreedhar Acharyya’s formula in a quadratic equation we get,
\(x=\frac{-2 \pm \sqrt{(2) 2-4.5(-7)}}{2 \times 5}\) \(x=\frac{-2 \pm \sqrt{4+140}}{10}=\frac{-2 \pm 12}{10}\)∴ \(\frac{\mathrm{k} \pm 12}{10}=\frac{-2 \pm 12}{10}\)
∴ k = -2.
| Chapter Name | Solutions | ||||
| Chapter 1 Quadratic Equations in One Variable | Exercise 1 | Exercise 2 | Exercise 3 | Exercise 4 | Exercise 5 |
| Chapter 2 Simple Interest | Exercise 1 | Exercise 2 | |||
| Chapter 3 Theorems related to Circle | Exercise 1 | Exercise 2 | |||
| Chapter 4 Rectangular Parallelopiped or Cuboid | Exercise 1 | Exercise 2 | |||
| Chapter 5 Ratio and Proportion | Exercise 1 | Exercise 2 | Exercise 3 | ||
| Chapter 6 Compound Interest and Uniform Rate of Increase or Decrease | Exercise 1 | Exercise 2 | |||
| Chapter 7 Theorems Related To Angles In A Circle | Exercise 1 | Exercise 2 | Exercise 3 | Exercise 4 | Exercise 5 |
| Chapter 8 Right Circular Cylinder | Exercise 1 | Exercise 2 | |||
| Chapter 9 Quadratic Surd | Exercise 1 | Exercise 2 | Exercise 3 | ||
| Chapter 10 Theorems Related To Cyclic Quadrilateral | Exercise 1 | ||||
| Chapter 11 Construction Of Circumcircle and Incircle Of A Traingle | Exercise 1 | Exercise 2 | |||
| Chapter 12 Sphere | Exercise 1 | Exercise 2 | |||
| Chapter 13 Variation | Exercise 1 | Exercise 2 | |||
| Chapter 14 Partnership Business | Exercise 1 | ||||
| Chapter 15 Theorems Related To Tangent To A Circle | Exercise 1 | Exercise 2 | |||
| Chapter 16 Right Circular Cone | Exercise 1 | ||||
| Chapter 17 Construction Of Tangent To A Circle | Exercise 1 | ||||
| Chapter 18 Similarity | Exercise 1 | Exercise 2 | Exercise 3 | Exercise 4 | |
| Chapter 19 Problems Related To Different Solid Objects | Exercise 1 | ||||
| Chapter 20 Trigonometry: Concept of Measurement Of Angle | Exercise 1 | ||||
| Chapter 21 Construction: Determination Of Mean Proportional | Exercise 1 | ||||
| Chapter 22 Pythagoras Theorem | Exercise 1 | ||||
| Chapter 23 Trigonometric Ratios And Trigonometric Identities | Exercise 1 | Exercise 2 | Exercise 3 | ||
| Chapter 24 Trigonometric Ratios Of Complementary Angle | Exercise 1 | ||||
| Chapter 25 Application Of Trigonometric Rations: Heights & Distances | Exercise 1 | ||||
| Chapter 26 Statistics: Mean, Median, Ogive, Mode | Exercise 1 | Exercise 2 | Exercise 3 | Exercise 4 |
Question 4. If the following quadratic equations have real roots, then let us determine them with the help of Sreedhar Acharyya’s formula.
1. 3x2+11x-4=0
Solution: 3x2 + 11x – 4 = 0
\(x=\frac{-11 \pm \sqrt{(11)^2-4.3(-4)}}{2 \times 3}\)= \(\frac{-11 \pm \sqrt{121+48}}{2 \times 3}\)
= \(\frac{-11 \pm \sqrt{169}}{6}=\frac{-11 \pm 13}{6}\)
∴ \(x=\frac{-11+13}{6}\)
& \(x=\frac{-11-13}{6}\)
x = \(\frac{2}{6}\)
x = \(\frac{1}{3}\)
& x = –\(\frac{24}{6}\)
x = -4
∴ x = 1/3 and x = -4.
2. (x-2) (x+4) +9=0
Solution: (x – 2)(x + 4) + 9 = 0
⇒ x2 + 2x – 8 + 9 = 0
⇒ x2 + 2x + 1 = 0
\(x=\frac{-(+2) \pm \sqrt{(+2)^2-4.1 .1}}{2 \times 1}=\frac{-2 \pm \sqrt{4-4}}{2}=\frac{-2 \pm 0}{2}\)∴ x = -1 & x = -1.
Question 5. Let us express the following mathematical problems in the form of quadratic equations with one variable and solve them by applying Sreedhar Acharyya’s formula or with the help of factorization.
1. Sathi has drawn a right-angled triangle whose length of the hypotenuse is 6 cm more than twice 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 the three sides of the right-angled triangle drawn by Sathi.
Solution: Let the length of smallest side of the triangle = x cm.
∴ Length of hypotenuse = (2x + 6)cm.
& length of the 3rd side = 2x + 6 – 2 = 2x + 4
∴ \((2 x+6)^2=x^2+(2 x+4)^2\)
⇒ \(4 x^2+24 x+36=x^2+4 x^2+16 x+16\)
⇒ \(x^2+16 x-24 x+16-36=0\)
⇒ \(x^2-8 x-20=0\)
∴ \(x=\frac{-(-8) \pm \sqrt{(-8)^2-4.1(-20)}}{2 \times 1}=\frac{8 \pm \sqrt{64+80}}{2}=\frac{8 \pm 12}{2}\)
∴ \(x=\frac{8+12}{2} \& x=\frac{8-12}{2}=-2\) (not possible)
⇒ \(x=\frac{20}{2}=10\)
∴ The length of the sides are 10cm, 24 cm & 26 cm.
2. If a two-digit positive number is multiplied by its unit digit, then the product is 189 and if the ten’s digit is twice of the unit digit, then let us calculate the unit digit.
Solution: Let in a two digit number the digit in the tenth place 2x.
∴ The number is 10(2x) + x = 21x.
According to the problem,
21x x = 189
⇒ 21x2 = 189
x2 = 9
∴ x = ±3
∴ The digit in the unit place = 3.
3. The speed of Salma is 1m/second more than the speed of Anik. In a 180 m run, Salma reaches 2 seconds before Anik. Let us write by calculating the speed of Anik in m/sec.
Solution: Let the speed of Anik is xm/sec.
∴ Speed of Salma is (x + 1)m/sec.
To go 180 m, Anik takes\(\frac{180}{x}\) sec & Salma takes \(\frac{180}{x+1}\)sec.
According to the problem,
\(\frac{180}{x}-\frac{180}{x+1}=2\)⇒ \(180\left(\frac{1}{x}-\frac{1}{x+1}\right)=2\)
⇒ \(\frac{x+1-x}{x(x+1)}=\frac{2}{180}=\frac{1}{90}\)
⇒ x2 + x = 90
⇒ x2 + x – 90 = 0
⇒ x2 + 10x – 90 – 90 = 0
⇒ x(x + 10) – 9(x + 10) = 0
⇒ (x + 10)(x – 9) = 0
∴ x = -10 (It is impossible)
or, x – 9 = 0
∴ x = 9
∴ Speed of Anis is 9m/sec
4. There is a square park in our locality. The area of a rectangular park is 78 sqm less than the twice of area of that square-shaped park whose length is 5 m more than the length of the side of that park and the breadth is 3 m 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.
Solution: Let each side of the square park = xm & its area = x2 sqm.
∴ Length of the rectangle = (x + 5)m
& the breadth of the rectangle = (x – 3)m.
Area of the rectangle = (x + 5)(x – 3)sqm.
According to the problem,
2x2 – (x + 5).(x – 3) = 78
⇒ 2x2 – (x2 + 5x – 3x – 15) – 78 = 0
⇒ 2x2 – x2 – 2x + 15 – 78 = 0
⇒ x2 – 2x – 63 = 0
⇒ x2 – 9x + 7x – 63 = 0
⇒ x(x – 9) + 7(x – 9) = 0
⇒ (x-9)(x+7)=0
∴ Eitherx-9=0 ∴x=9.
Or, x+7=0 ∴x=-7 (Not possible)
∴ Length of each side of the square field=9 m. Ans.
5. In our village, Proloy babu bought 350 chili plants for planting on his rectangular land. When he put the plants in rows, he noticed that if he would put 24 rows more than the number of plants in each row, 10 plants would remain in excess. Let us write by calculating the number of plants he put in each row.
Solution: Let proloy baby puts x plants in each row.
According to the problem,
x x (x + 24) + 10 = 350
⇒ x2 + 24x + 10 – 350 = 0
⇒ x2 + 24x – 340 = 0
⇒ x2 + (34 – 10)x – 340 = 0
⇒ x2 + 34x – 10x – 340 = 0
⇒ x(x + 34) – 10(x + 34) = 0
⇒ (x + 34)(x – 10) = 0
Either x + 34 = 0
∴ x = -34 (It is impossible)
or, x – 10 = 0
∴ x = 10
∴ No. of plants in each row = 10.
6. 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.
Solution: Let Kuntal takes x min. to make an article & Joesph takes (x – 5)min to make the article.
∴ In 6 hrs. Kuntal makes \(\frac{6 \times 60}{x}\) & Joesph makes \(\frac{6 \times 60}{x-5}\) articles respectively.
∴ According to the problem,
\(\frac{6 \times 60}{x-5}-\frac{6 \times 60}{x}=6\)or, \(6 \times 60\left(\frac{1}{x-5}-\frac{1}{60}\right)=6\)
or, \(\frac{x-x+5}{x(x-5)}=\frac{1}{60}\)
∴ x2 – 5x = 300
or, x2 – 5x – 300 = 0
or, x2 – 20x + 15x – 300 = 0
or, x(x – 20) + 15(x – 20) = 0
or, (x – 20)(x + 15) = 0
either x – 20 = 0
or, x + 15 = 0
∴ x = 20
∴ in 20 min. kuntal makes 1 article.
∴ In 6hrs = 6 x 60 min.
Kuntal makes \(\frac{1}{20} \times 6 \times 60=18\) articles.
7. The speed of a boat in still water is 8 km/hr. If the boat can go 15 km downstream and 22 km upstream in 5 hours, then let us write by calculating, the speed of the stream.
Solution: Let the speed of current = x km/hr
As the speed of the boat = 8 km/hr
∴ Speed of the boat with the current = (8 + x)km/hr
& Speed of the boat against the current = (8 – x)km/hr
∴ According to the problem,
\(\frac{15}{8+x}+\frac{22}{8-x}=5\)or, \(\frac{15(8-x)+22(8+x)}{(8+x)(8-x)}=5\)
or, 120 – 15x + 176 + 22x = 5(64 – x2)
or, 296 + 7x = 320 – 5x2
or, 5x2 + 7x – 24 = 0
or, 5x2 + 15x – 8x – 24 = 0
or, 5x(x + 3) -8(x + 3) = 0
or, (x + 3).(5x – 8) = 0
either x + 3 = 0
x = -3 (not possible)
or, 5x – 8 = 0
5x = 8
∴ \(x=\frac{8}{5}=1 \frac{3}{5}\)
∴ Speed of the current = \(1 \frac{3}{5}\) km/hr
8. A superfast train runs having a speed of 15 km/hr more than that of an express train. Leaving the same station the superfast train reached a station of 180 km distance 1 hour before that the express train. Let us determine the speed of the superfast train in km/hr.
Solution: Let the speed of the superfast train = \(\frac{x \mathrm{~km}}{\mathrm{hr}}\) &
the speed of the express train = (x – 15)km/hr
According to the problem,
\(\frac{180}{x-15}-\frac{180}{x}=1\)or, \(180\left(\frac{1}{x-15}-\frac{1}{x}\right)=1\)
or, \(\frac{x-x+15}{x(x-15)}=\frac{1}{180}\)
or, x2 – 15x = 15 x 180
or, x2 – 15x – 2700 = 0
or, x2 – 60x + 45x – 2700 = 0
or, x(x – 650) + 45(x – 60) = 0
or, (x – 60)(x + 45) = 0
Either x – 60 = 0
x = 60
or, x + 45 = 0
x = -45 (not possible)
∴ Speed of the superfast train = 60km/hr.
9. 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 fish of 1 kg. The quantity of fish and that of dal in Rs. 240 is equal to the quantity of rice in Rs. 280. Let us calculate the cost price of 1 kg of fish.
Solution: Let the price of fish = Rs x /kg &
the price of rice = Rs. (x – 40)/kg &
the price of pulse = Rs (x – 20)/kg
∴ According to the problem,
\(\frac{240}{x}+\frac{240}{x-20}=\frac{280}{x-40}\) \(240\left[\frac{1}{x}+\frac{1}{x-20}\right]=\frac{280}{x-40}\) \(6\left[\frac{x-20+x}{x(x-20)}\right]=\frac{7}{x-40}\) \(6 \times \frac{(2 x-20)}{x^2-20 x}=\frac{7}{x-40}\)or, 6 x (2x – 20)(x – 40) = 7(x2 – 20x)
or, 12x2 – 7x2 – 600x + 140x + 4800 – 0
or, 5x2 – 460x + 4800 = 0
or, 5(x2 – 92x + 960) = 0
x2 – 80x – 12x + 960 = 0
x(x – 80) – 12(x – 80) = 0
or, (x – 80)(x – 12) = 0
Either x – 80 = 0
∴ x = 80
or, x – 12 = 0
∴ x = 12 (Not possible)
∴ Price of Fish = Rs. 80/kg
Question 6. I determine the nature of the two roots of the following quadratic equations: 2x2+x-2
Solution: 2x2 + x – 2 = 0
Here, Discriminant = b2 – 4ac
= (1)2 – 4.2(-2)
= 1 + 16 = 17 > 0
∴ Roots are real & unequal.
Question 7. By understanding, let us write the value of k for which the two roots of the quadratic equation 2x2 – 10x + k = 0 are real and equal.
Solution: 2x2 – 10x + k = 0
Here, Discriminant = b2 – 4ac
= (-10)2 – 4.2.k
= 100 – 8k
As the roots of the equation are real & equal,
∴ 100 – 8k = 0
∴ 8k = 100
∴ \(k=\frac{100}{8}=\frac{25}{2}\)
Question 8. I determine the sum and product of two roots of the quadratic equations: 4x2-9x=100
Solution: 4x2-9x-100 = 0
Sum of roots = \(-\frac{b}{a}=\frac{-(-9)}{4}=\frac{9}{4}\)
Product of roots = \(\frac{c}{a}=\frac{-100}{4}=-25\)
Question 9. If one of the roots of the quadratic equation 3x2 – 10x + 3 = 0 is then let me 3. determine the other root of it.
Solution: 3x2 – 10x + 3 = 0
If one root = \(\frac{1}{3}\) & let the other root = x.
∴ Sum of the roots = \(\frac{-(-10)}{3}=\frac{10}{3}\)
\(\frac{1}{3}+x=\frac{10}{3}\)∴ \(x=\frac{10}{3}-\frac{1}{3}=\frac{9}{3}=3\)
Question 10. If a and ẞ are two roots of the quadratic equation ax2 + bx + c = 0 [a + 0], then let Us express the value of (1/α2 + 1/ ẞ2) in terms of a, b, and c.
Solution: Given, ax2 + bx + c = 0. [x and β are the roots of the equation]
∴ \(\alpha+\beta=\frac{-\mathrm{b}}{\mathrm{a}} \& \propto \beta=\frac{\mathrm{c}}{\mathrm{a}}\)
\(\frac{1}{\alpha^3}+\frac{1}{\beta^3}=\left(\frac{\alpha^3+\beta^3}{\alpha^3 \beta^3}\right)=\frac{(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)}{(\alpha \beta)^3}=\frac{\left(\frac{-b}{a}\right)^3-3 \frac{c}{a}\left(-\frac{b}{a}\right)}{(c / a)^3}\)= \(\frac{\frac{-b^3}{a^3}+\frac{3 b c}{a^2}}{c^3 / a^3}=\frac{\frac{-b^3+3 a b c}{a^3}}{c^3 / a^3}=\frac{3 a b c-b^3}{c^3}\)
Question 11. By determining, we are observing that two roots of the quadratic equation x2-7x+12= 0 are 3 and 4.
Solution: x2 – 7x + 12 = 0
∴ \(x=\frac{-(-7) \pm \sqrt{(-7)^2-4.1 .12}}{2 \times 1}=\frac{7 \pm 1}{2}\)
∴ \(x=\frac{7+1}{2}=4\)
\(x=\frac{7-1}{2}=3\)