NCERT Class 10 Maths Exercise 4.2 Solution – Chapter 4

Solution:

1. $$x^2-3x-10=0$$

   We are given x² – 3x – 10 = 0, and we have to solve it by the factorisation method.

2. $$\text{Find two numbers: product}=-10,\text{sum}=-3$$

   For factorisation, we need 2 numbers whose product is -10 and whose sum is -3.

3. $$(-5)\times (+2)=-10,(-5)+(+2)=-3$$

   The two numbers are -5 and +2, because their product is -10 and their sum is -3.

4. $$x^2-5x+2x-10=0$$

   Now split the middle term -3x into -5x + 2x.

5. $$x(x-5)+2(x-5)=0$$

   Now take out common factors. Take x from the first two terms and 2 from the last two terms.

6. $$(x-5)(x+2)=0$$

   Now x – 5 is a common factor, so we get (x – 5)(x + 2) = 0.

7. $$x-5=0\Rightarrow x=5$$

   Set the first factor equal to zero: x – 5 = 0, so x = 5.

8. $$x+2=0\Rightarrow x=-2\therefore x=5\text{ or }x=-2$$

   Set the second factor equal to zero: x + 2 = 0, so x = -2. So the two roots are x = 5 and x = -2.

Solution:

1. $$2x^2+x-6=0$$

   Our equation is 2x² + x – 6 = 0. We have to solve it by the factorisation method.

2. $$a\times c=2\times (-6)=-12,b=+1$$

   For factorisation, we must find 2 numbers whose product is -12 and whose sum is +1.

3. $$(-3)\times 4=-12\text{and}(-3)+4=+1\mathrm{✓}$$

   The 2 numbers are -3 and +4, because their product is -12 and their sum is +1.

4. $$2x^2+4x-3x-6=0$$

   Now split the middle term x into -3x + 4x and write the equation.

5. $$2x(x+2)-3(x+2)=0$$

   Take 2x from the first 2 terms and -3 from the next 2 terms.

6. $$(2x-3)(x+2)=0$$

   Now take out the common factor x + 2. We get (2x – 3)(x + 2) = 0.

7. $$2x-3=0\Rightarrow x={\displaystyle \frac{3}{2}}\text{or}x+2=0\Rightarrow x=-2$$

   By the zero product property, either 2x – 3 = 0, which gives x = 3/2, or x + 2 = 0, which gives x = -2.

Solution:

1. $$\sqrt{2}{x}^2+7x+5\sqrt{2}=0$$

   Our equation is √2x² + 7x + 5√2 = 0. We have to solve it by the factorisation method.

2. $$a\cdot c=\sqrt{2}\times 5\sqrt{2}=5\times 2=10,b=7$$

   For factorisation, we must split the middle term 7x. For this, find a × c, which is √2 × 5√2 = 10. We need 2 numbers whose product is 10 and whose sum is 7.

3. $$2+5=7\text{and}2\times 5=10\mathrm{✓}$$

   The 2 numbers are 2 and 5, because 2 + 5 = 7 and 2 × 5 = 10. Perfect!

4. $$\sqrt{2}{x}^2+5x+2x+5\sqrt{2}=0$$

   Now split the middle term 7x into 5x + 2x and rewrite the equation.

5. $$x(\sqrt{2}x+5)+\sqrt{2}(\sqrt{2}x+5)=0$$

   Take x from the first 2 terms to get x(√2x + 5). Take √2 from the next 2 terms. Since √2 × √2 = 2, we get √2(√2x + 5).

6. $$(\sqrt{2}x+5)(x+\sqrt{2})=0$$

   Now take out the common factor √2x + 5. We get 2 factors: √2x + 5 and x + √2.

7. $$\sqrt{2}x+5=0\Rightarrow x={\displaystyle \frac{-5}{\sqrt{2}}}={\displaystyle \frac{-5\sqrt{2}}{2}}$$

   From the first factor: √2x = -5, so x = -5/√2. Rationalising the denominator gives x = -5√2/2.

8. $$x+\sqrt{2}=0\Rightarrow x=-\sqrt{2}\therefore x=-\sqrt{2},{\displaystyle \frac{-5\sqrt{2}}{2}}$$

   From the second factor: x = -√2. So our two roots are x = -√2 and x = -5√2/2.

Solution:

1. $$2x^2-x+\frac{1}{8}=0$$

   We are given 2x² – x + 1/8 = 0. We have to solve it by the factorisation method.

2. $$\text{Multiply both sides by }8:16x^2-8x+1=0$$

   First make the equation simple. Multiply both sides by 8 to remove the fraction. We get 16x² – 8x + 1 = 0.

3. $$16x^2-8x+1=0\Rightarrow 16x^2-4x-4x+1=0$$

   Now split the middle term -8x into -4x – 4x, because (-4x) + (-4x) = -8x and (-4x) × (-4x) = 16x² × 1.

4. $$4x(4x-1)-1(4x-1)=0$$

   Now group the terms. Take 4x from the first two terms and -1 from the last two. We get 4x(4x – 1) – 1(4x – 1) = 0.

5. $$(4x-1)(4x-1)=0\Rightarrow (4x-1{)}^2=0$$

   Both brackets are the same, so this is a perfect square: (4x – 1)² = 0.

6. $$4x-1=0\Rightarrow x=\frac{1}{4}$$

   From 4x – 1 = 0 we get x = 1/4. Since both roots are the same, they are called repeated or equal roots.

7. $$x=\frac{1}{4},x=\frac{1}{4}(\text{equal roots})$$

   So both roots of this quadratic equation are equal, and the answer is x = 1/4.

Solution:

1. $$100x^2-20x+1=0$$

   We are given this quadratic equation and we have to solve it by the factorisation method.

2. $$\text{Find 2 numbers: product}=100\times 1=100,\text{sum}=-20$$

   For factorisation, we need 2 numbers whose product is 100 and whose sum is -20.

3. $$-10\times (-10)=100,-10+(-10)=-20\mathrm{✓}$$

   The two numbers are -10 and -10, because their product is 100 and their sum is -20.

4. $$100x^2-10x-10x+1=0$$

   Now split the middle term -20x into -10x and -10x.

5. $$10x(10x-1)-1(10x-1)=0$$

   Now take 10x from the first 2 terms and -1 from the last 2 terms.

6. $$(10x-1)(10x-1)=0$$

   Now take out the common factor 10x – 1. We get (10x – 1)² = 0.

7. $$10x-1=0⟹x={\displaystyle \frac{1}{10}}$$

   Both factors are equal, so 10x – 1 = 0 gives x = 1/10.

8. $$x={\displaystyle \frac{1}{10}}\text{ and }x={\displaystyle \frac{1}{10}}(\text{equal roots})$$

   Both roots of this equation are equal: x = 1/10 and x = 1/10.

Solution:

1. $$\text{Let John’s marbles}=x,\text{Jivanti’s marbles}=45-x$$

   Let John have x marbles. Then Jivanti has 45 – x marbles, because together they have 45.

2. $$\text{After losing 5 each: John}=x-5,\text{Jivanti}=40-x$$

   Each of them lost 5 marbles. So John has x – 5 left and Jivanti has 45 – x – 5, which is 40 – x marbles.

3. $$(x-5)(40-x)=124$$

   The condition says the product of their marbles is 124. So we write (x – 5)(40 – x) = 124.

4. $$40x-x^2-200+5x=124$$

   Expand the left side. Opening the brackets gives 40x – x² – 200 + 5x = 124.

5. $$x^2-45x+324=0$$

   Move all the terms to one side and simplify. We get the standard quadratic equation x² – 45x + 324 = 0.

6. $$x^2-36x-9x+324=0⟹x(x-36)-9(x-36)=0$$

   Now factorise. Split the middle term -45x into -36x and -9x, then take out the common factor.

7. $$(x-9)(x-36)=0⟹x=9\text{ or }x=36$$

   So (x – 9)(x – 36) = 0, which means x = 9 or x = 36. Both are valid solutions.

8. $$\text{If }x=36:\text{John}=36,\text{Jivanti}=9\text{If }x=9:\text{John}=9,\text{Jivanti}=36$$

   When x = 36, John had 36 and Jivanti had 9 marbles. When x = 9, John had 9 and Jivanti had 36 marbles. Both cases give the same answer.

Solution:

1. $$\text{Total Revenue}=\text{Units}\times \text{Price per Unit}$$

   This idea is very useful in real life. In a factory, the more items you make, the lower the cost per item, and the total cost becomes a quadratic equation.

2. $$\text{Let number of toys}=x$$

   First, let the number of toys made that day be x.

3. $$\text{Cost per toy}=55-x$$

   The problem says the cost of each toy is 55 minus the number of toys made that day. So the cost per toy is Rs. (55 – x).

4. $$\text{Total Cost}=x\times (55-x)=750$$

   Total cost = number of toys × cost per toy, and this is given as Rs. 750. So the equation is x(55 – x) = 750.

5. $$55x-x^2=750$$

   Open the bracket to get 55x – x² = 750.

6. $$x^2-55x+750=0$$

   Move everything to one side: x² – 55x + 750 = 0. This is our quadratic equation.

7. $$x^2-25x-30x+750=0$$

   Now split the middle term. Break 55x into 25x and 30x, because 25 × 30 = 750 and 25 + 30 = 55.

8. $$x(x-25)-30(x-25)=0\Rightarrow (x-25)(x-30)=0$$

   Take out the common factor: x(x – 25) – 30(x – 25) = 0, so (x – 25)(x – 30) = 0.

9. $$x=25\text{or}x=30$$

   Both values are valid. At x = 25 the cost per toy is Rs. 30, and at x = 30 the cost per toy is Rs. 25. Both are positive, so the number of toys made is 25 or 30.

Solution:

1. $$\text{Let the two numbers be }x\text{ and }(27-x)$$

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   Let the first number be x. Then the second number is 27 – x, because their sum is 27.

2. $$x(27-x)=182$$

   The condition says the product of the two numbers is 182. So we write x(27 – x) = 182.

3. $$27x-x^2=182$$

   Expand the left side: 27x – x² = 182.

4. $$x^2-27x+182=0$$

   Move everything to one side: x² – 27x + 182 = 0. This is our quadratic equation.

5. $$x^2-13x-14x+182=0$$

   Now split the middle term. 13 × 14 = 182 and 13 + 14 = 27, so write -27x as -13x – 14x.

6. $$x(x-13)-14(x-13)=0\Rightarrow (x-13)(x-14)=0$$

   Take out the common factor (x – 13). We get (x – 13)(x – 14) = 0.

7. $$x=13\text{or}x=14$$

   So x = 13 or x = 14. Both are valid solutions.

8. $$\text{Numbers: }13\text{ and }14(13+14=27,13\times 14=182)$$

   The two numbers are 13 and 14. Check: 13 + 14 = 27 ✓ and 13 × 14 = 182 ✓.

Solution:

1. $$\text{Area of two square rooms}={13}^2+{14}^2=365\text{ sq. ft.}$$

   Think of real life: if 2 square rooms have sides that are consecutive integers and their total area is 365 square feet, we get this same equation.

2. $$\text{Let integers}=n\text{ and }(n+1)$$

   Let the two consecutive positive integers be n and n + 1.

3. $$n^2+(n+1{)}^2=365$$

   The sum of their squares is 365, so the equation is n² + (n + 1)² = 365.

4. $$n^2+n^2+2n+1=365$$

   Expand (n + 1)² to get n² + 2n + 1. So the left side becomes 2n² + 2n + 1.

5. $$2n^2+2n+1-365=0\Rightarrow 2n^2+2n-364=0$$

   Move 365 to the left side. The equation becomes 2n² + 2n – 364 = 0.

6. $$n^2+n-182=0$$

   Divide the whole equation by 2. It simplifies to n² + n – 182 = 0.

7. $$n^2+14n-13n-182=0\Rightarrow n(n+14)-13(n+14)=0$$

   Split the middle term using 14 and 13, then factorise: n(n + 14) – 13(n + 14) = 0.

8. $$(n-13)(n+14)=0\Rightarrow n=13\text{ or }n=-14$$

   The factors are n – 13 and n + 14, so n = 13 or n = -14. But the integers must be positive, so we reject n = -14.

9. $$n=13,n+1=14\Rightarrow {13}^2+{14}^2=169+196=365\mathrm{✓}$$

   So the two consecutive positive integers are 13 and 14. Check: 169 + 196 = 365, which is correct!

Solution:

1. $$\text{Ladder against wall: base, height, hypotenuse}=13\text{ cm}$$

   This is just like a ladder leaning against a wall, where the base, the height and the ladder form a right triangle.

2. $$\text{Let base}=x\text{ cm, altitude}=(x-7)\text{ cm}$$

   Let the base of the right triangle be x cm. The altitude is 7 cm less than the base, so the altitude is x – 7 cm.

3. $$\text{Hypotenuse}=13\text{ cm (given)}$$

   The hypotenuse is given as 13 cm, and we will use the Pythagoras theorem.

4. $$x^2+(x-7{)}^2={13}^2$$

   By the Pythagoras theorem, base² + altitude² = hypotenuse². So x² + (x – 7)² = 13².

5. $$x^2+x^2-14x+49=169$$

   Now expand (x – 7)² to get x² – 14x + 49, and 13² = 169.

6. $$2x^2-14x+49-169=0\Rightarrow 2x^2-14x-120=0$$

   Add the two x² terms and move 169 to the left side. We get 2x² – 14x – 120 = 0.

7. $$x^2-7x-60=0$$

   Divide the whole equation by 2. It simplifies to x² – 7x – 60 = 0.

8. $$x^2-12x+5x-60=0\Rightarrow x(x-12)+5(x-12)=0\Rightarrow (x+5)(x-12)=0$$

   Now split the middle term -7x into -12x + 5x and factorise. We get (x + 5)(x – 12) = 0.

9. $$x=12(\text{since }x=-5\text{ rejected, side}>0)$$

   So x = 12 or x = -5. But the length of a side cannot be negative, so we reject x = -5. The base is 12 cm and the altitude is 12 – 7 = 5 cm.

Solution:

1. $$\text{Total cost}=\text{(no. of items)}\times \text{(price per item)}$$

   This idea is used everywhere in real life. In a shop, total bill = quantity × price per item. When the price depends on the quantity, the equation becomes quadratic.

2. $$\text{Let number of articles}=x$$

   First, set the variable. Let the number of articles made that day be x.

3. $$\text{Cost per article}=2x+3\text{ (Rs)}$$

   The question says the cost of each article is 3 more than twice the number of articles made that day. So the cost per article is Rs. (2x + 3).

4. $$\text{Total cost}=x\times (2x+3)=90$$

   Total cost = number of articles × cost per article, and this is given as Rs. 90. So the equation is x(2x + 3) = 90.

5. $$2x^2+3x-90=0$$

   Open the bracket to get 2x² + 3x = 90. Subtract 90 from both sides to get 2x² + 3x – 90 = 0.

6. $$2x^2+15x-12x-90=0$$

   Now solve this by factorisation. Split the middle term 3x so that the two numbers have product 2 × (-90) = -180 and sum 3. The numbers are 15 and -12.

7. $$x(2x+15)-6(2x+15)=0\Rightarrow (2x+15)(x-6)=0$$

   Take out the common factor: x(2x + 15) – 6(2x + 15) = 0, so the factors are (2x + 15)(x – 6) = 0.

8. $$x=-\frac{15}{2}\text{ (rejected)},x=6$$

   From 2x + 15 = 0 we get x = -15/2, but the number of articles cannot be negative or a fraction, so we reject it. From x – 6 = 0 we get x = 6, which is the valid answer.

9. $$x=6,\text{Cost per article}=2(6)+3=\text{Rs }15$$

   So 6 articles were made that day, and the cost of each article is 2 × 6 + 3 = Rs. 15. This is our final answer.