NCERT Solutions Class 10 Maths Chapter 4: Quadratic Equations

#### **Q1. Find the roots of the following quadratic equations by factorisation:** **(i)** $x^2 – 3x – 10 = 0$ **(ii)** $2x^2 + x – 6 = 0$ **(iii)** $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$ **(iv)** $2x^2 – x + \frac{1}{8} = 0$ **(v)** $100x^2 – 20x + 1 = 0$ #### **A1. Solution:** **(i) $x^2 – 3x – 10 = 0$** Split the middle term $-3x$ into $-5x + 2x$: $x^2 - 5x + 2x - 10 = 0$ $x(x - 5) + 2(x - 5) = 0$ $(x - 5)(x + 2) = 0$ $x - 5 = 0 \Rightarrow x = 5$ $x + 2 = 0 \Rightarrow x = -2$ **Roots are $5$ and $-2$.** **(ii) $2x^2 + x – 6 = 0$** Split $x$ into $4x - 3x$: $2x^2 + 4x - 3x - 6 = 0$ $2x(x + 2) - 3(x + 2) = 0$ $(x + 2)(2x - 3) = 0$ $x + 2 = 0 \Rightarrow x = -2$ $2x - 3 = 0 \Rightarrow x = \frac{3}{2}$ **Roots are $-2$ and $\frac{3}{2}$.** **(iii) $\sqrt{2}x^2 + 7x + 5\sqrt{2} = 0$** Split $7x$ into $5x + 2x$: $\sqrt{2}x^2 + 5x + 2x + 5\sqrt{2} = 0$ $x(\sqrt{2}x + 5) + \sqrt{2}(\sqrt{2}x + 5) = 0$ *(Note: $2x = \sqrt{2} \cdot \sqrt{2}x$)* $(\sqrt{2}x + 5)(x + \sqrt{2}) = 0$ $\sqrt{2}x + 5 = 0 \Rightarrow x = -\frac{5}{\sqrt{2}}$ $x + \sqrt{2} = 0 \Rightarrow x = -\sqrt{2}$ **Roots are $-\frac{5}{\sqrt{2}}$ and $-\sqrt{2}$.** **(iv) $2x^2 – x + \frac{1}{8} = 0$** Multiply by 8 to remove the fraction: $16x^2 - 8x + 1 = 0$ Split $-8x$ into $-4x - 4x$: $16x^2 - 4x - 4x + 1 = 0$ $4x(4x - 1) - 1(4x - 1) = 0$ $(4x - 1)(4x - 1) = 0$ $4x - 1 = 0 \Rightarrow x = \frac{1}{4}$ **Roots are $\frac{1}{4}$ and $\frac{1}{4}$.** **(v) $100x^2 – 20x + 1 = 0$** Split $-20x$ into $-10x - 10x$: $100x^2 - 10x - 10x + 1 = 0$ $10x(10x - 1) - 1(10x - 1) = 0$ $(10x - 1)(10x - 1) = 0$ $10x - 1 = 0 \Rightarrow x = \frac{1}{10}$ **Roots are $\frac{1}{10}$ and $\frac{1}{10}$.** #### **Q2. Solve the problems given in Example 1.** #### **A2. Solution:** **(i) Problem Statement:** 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. **Solution:** Let John's marbles = $x$. Then Jivanti's marbles = $45 - x$. After losing 5 marbles: John = $x - 5$ Jivanti = $(45 - x) - 5 = 40 - x$ Product = 124 $(x - 5)(40 - x) = 124$ $40x - x^2 - 200 + 5x = 124$ $-x^2 + 45x - 200 - 124 = 0$ $-x^2 + 45x - 324 = 0$ $x^2 - 45x + 324 = 0$ **Factorisation:** Find two numbers with sum 45 and product 324. (36 and 9) $x^2 - 36x - 9x + 324 = 0$ $x(x - 36) - 9(x - 36) = 0$ $(x - 36)(x - 9) = 0$ $x = 36$ or $x = 9$ **Answer: If John had 36, Jivanti had 9. If John had 9, Jivanti had 36.** **(ii) Problem Statement:** 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 ₹ 750. We would like to find out the number of toys produced on that day. **Solution:** Let the number of toys produced = $x$. Cost of each toy = $55 - x$. Total cost = $x(55 - x) = 750$. $55x - x^2 = 750$ $x^2 - 55x + 750 = 0$ **Factorisation:** Find two numbers with sum 55 and product 750. (30 and 25) $x^2 - 30x - 25x + 750 = 0$ $x(x - 30) - 25(x - 30) = 0$ $(x - 30)(x - 25) = 0$ $x = 30$ or $x = 25$ **Answer: The number of toys produced is either 30 or 25.** #### **Q3. Find two numbers whose sum is 27 and product is 182.** #### **A3. Solution:** Let the first number be $x$. Then the second number is $27 - x$. Product = $182$ $x(27 - x) = 182$ $27x - x^2 = 182$ $x^2 - 27x + 182 = 0$ **Factorisation:** Find two numbers with sum 27 and product 182. (14 and 13) $x^2 - 14x - 13x + 182 = 0$ $x(x - 14) - 13(x - 14) = 0$ $(x - 14)(x - 13) = 0$ $x = 14$ or $x = 13$ **Answer: The two numbers are 13 and 14.** #### **Q4. Find two consecutive positive integers, sum of whose squares is 365.** #### **A4. Solution:** Let the consecutive integers be $x$ and $x+1$. $x^2 + (x + 1)^2 = 365$ $x^2 + (x^2 + 1 + 2x) = 365$ $2x^2 + 2x + 1 - 365 = 0$ $2x^2 + 2x - 364 = 0$ Divide by 2: $x^2 + x - 182 = 0$ **Factorisation:** Find two numbers with difference 1 and product 182. (14 and 13) $x^2 + 14x - 13x - 182 = 0$ $x(x + 14) - 13(x + 14) = 0$ $(x + 14)(x - 13) = 0$ $x = -14$ or $x = 13$ Since integers are positive, we reject $-14$. So, $x = 13$. Next integer $x + 1 = 14$. **Answer: The integers are 13 and 14.** #### **Q5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.** #### **A5. Solution:** Let the Base = $x$ cm. Then Altitude = $x - 7$ cm. By Pythagoras theorem: $(\text{Base})^2 + (\text{Altitude})^2 = (\text{Hypotenuse})^2$ $x^2 + (x - 7)^2 = (13)^2$ $x^2 + (x^2 + 49 - 14x) = 169$ $2x^2 - 14x + 49 - 169 = 0$ $2x^2 - 14x - 120 = 0$ Divide by 2: $x^2 - 7x - 60 = 0$ **Factorisation:** Find factors of 60 with difference 7. (12 and 5) $x^2 - 12x + 5x - 60 = 0$ $x(x - 12) + 5(x - 12) = 0$ $(x - 12)(x + 5) = 0$ $x = 12$ or $x = -5$ Length cannot be negative, so $x = 12$. Base = 12 cm. Altitude = $12 - 7 = 5$ cm. **Answer: Base = 12 cm, Altitude = 5 cm.** #### **Q6. 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 ₹ 90, find the number of articles produced and the cost of each article.** #### **A6. Solution:** Let the number of articles produced = $x$. Cost of each article = $2x + 3$. Total cost = $90$. $x(2x + 3) = 90$ $2x^2 + 3x - 90 = 0$ **Factorisation:** Product needed = $2 \times -90 = -180$. Sum needed = $3$. Factors are $15$ and $-12$. $2x^2 + 15x - 12x - 90 = 0$ $x(2x + 15) - 6(2x + 15) = 0$ $(2x + 15)(x - 6) = 0$ $x = 6$ or $x = -\frac{15}{2}$ Number of articles cannot be negative/fractional. So, $x = 6$. Cost of each article = $2(6) + 3 = 12 + 3 = 15$. **Answer: Number of articles = 6, Cost of each article = ₹15.**