NCERT Class 10 Maths Exercise 4.1 Solution – Chapter 4

Solution:

1. $$\text{Given: }(x+1{)}^2=2(x-3)$$

   We have to check whether this equation is quadratic or not. For this, we will write it in standard form.

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

   On the left side, expand (x + 1)² to get x² + 2x + 1. On the right side, multiply the bracket by 2 to get 2x – 6.

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

   Now move all the terms from the right side to the left side. The signs will change.

4. $$x^2+0x+7=0\Rightarrow x^2+7=0$$

   2x – 2x cancels, and 1 + 6 = 7. So we get x² + 7 = 0.

5. $$\text{Degree}=2\Rightarrow \text{It IS a Quadratic Equation}$$

   The highest power of x in this equation is 2, so its degree is 2. So this is a quadratic equation. It is in the standard form ax² + bx + c = 0, where a = 1, b = 0 and c = 7.

Solution:

1. $$x^2-2x=(-2)(3-x)$$

   We have to check whether this equation is quadratic or not. First, expand the right side.

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

   Multiply the bracket by -2. So -2 × 3 = -6 and -2 × (-x) = +2x.

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

   Now move all the terms to the left side. The right side becomes zero.

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

   Combine like terms: -2x – 2x = -4x. So the equation becomes x² – 4x + 6 = 0.

5. $$\text{Standard form: }a{x}^2+bx+c=0,a\mathrm{\ne }0$$

   The standard form of a quadratic equation is ax² + bx + c = 0, where a must not be zero.

6. $$a=1,b=-4,c=6,a\mathrm{\ne }0$$

   In our equation, a = 1, b = -4, c = 6, and a is not zero.

7. $$\therefore x^2-4x+6=0\text{ is a quadratic equation}$$

   So this is a quadratic equation, because it can be written in the standard form ax² + bx + c = 0 where a is not zero.

Solution:

1. $$\text{Given: }(x-2)(x+1)=(x-1)(x+3)$$

   We have to check whether this equation is quadratic or not. First, expand both sides.

2. $$\text{LHS: }(x-2)(x+1)=x^2+x-2x-2=x^2-x-2$$

   Expand the left side. x × x = x², then x – 2x = -x, and -2 × 1 = -2.

3. $$\text{RHS: }(x-1)(x+3)=x^2+3x-x-3=x^2+2x-3$$

   Now expand the right side. x × x = x², 3x – x = 2x, and -1 × 3 = -3.

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

   Now write LHS = RHS. Both sides have x².

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

   Move all the terms to the left side, that is, move the RHS across to the left.

6. $$-3x+1=0\Rightarrow -3x+1=0$$

   x² – x² = 0, -x – 2x = -3x, and -2 + 3 = 1. So we get -3x + 1 = 0.

7. $$\text{Degree}=1\mathrm{\ne }2\Rightarrow \text{NOT a Quadratic Equation}$$

   The highest power of x in this equation is 1, not 2. A quadratic equation must have degree 2, so this is not a quadratic equation.

Solution:

1. $$\text{Given: }(x-3)(2x+1)=x(x+5)$$

   We have to check whether this equation is quadratic or not. First, expand the left side.

2. $$\text{LHS: }(x-3)(2x+1)=2x^2+x-6x-3=2x^2-5x-3$$

   On the left side, multiply (x – 3) and (2x + 1). x × 2x = 2x², x × 1 = x, -3 × 2x = -6x, and -3 × 1 = -3. So the LHS is 2x² – 5x – 3.

3. $$\text{RHS: }x(x+5)=x^2+5x$$

   Now expand the right side. x(x + 5) = x² + 5x. This is very simple.

4. $$2x^2-5x-3=x^2+5x$$

   Now write LHS = RHS. Put both sides together.

5. $$2x^2-x^2-5x-5x-3=0\Rightarrow x^2-10x-3=0$$

   Now move all the terms to the left side. 2x² – x² = x², -5x – 5x = -10x, and -3 stays. So the equation is x² – 10x – 3 = 0.

6. $$\text{Degree}=2\Rightarrow \text{It IS a Quadratic Equation}$$

   The highest power of x in this equation is 2, so its degree is 2. In the standard form ax² + bx + c = 0, here a = 1, b = -10, c = -3 and a ≠ 0. So this is a quadratic equation.

Solution:

1. $$\text{Given: }(2x-1)(x-3)=(x+5)(x-1)$$

   We have to check whether this equation is quadratic or not. For this, we will expand both sides.

2. $$\text{LHS: }(2x-1)(x-3)=2x^2-6x-x+3=2x^2-7x+3$$

   Expand the left side: 2x × x = 2x², 2x × (-3) = -6x, -1 × x = -x, and -1 × (-3) = +3. Combining gives 2x² – 7x + 3.

3. $$\text{RHS: }(x+5)(x-1)=x^2-x+5x-5=x^2+4x-5$$

   Now expand the right side: x × x = x², x × (-1) = -x, 5 × x = 5x, and 5 × (-1) = -5. Combining gives x² + 4x – 5.

4. $$2x^2-7x+3=x^2+4x-5$$

   Now write LHS = RHS and move all the terms to one side.

5. $$2x^2-x^2-7x-4x+3+5=0$$

   Move all the RHS terms to the LHS, that is, find the difference of the two sides.

6. $$x^2-11x+8=0$$

   Simplifying gives x² – 11x + 8 = 0. This is the standard form.

7. $$\text{Degree}=2\Rightarrow \text{Quadratic Equation}\mathrm{✓}$$

   The highest power of x in this equation is 2, so its degree is 2. So this is a quadratic equation.

Solution:

1. $$x^2+3x+1=(x-2{)}^2$$

   We have to check whether this equation is quadratic or not. For this, we will expand the right side.

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

   On the right side, use the formula for (a – b)². So (x – 2)² = x² – 4x + 4.

3. $$x^2+3x+1=x^2-4x+4$$

   Now write the expanded form of the right side in the equation.

4. $$x^2+3x+1-x^2+4x-4=0$$

   Now move all the right side terms to the left side. The signs will change.

5. $$(x^2-x^2)+(3x+4x)+(1-4)=0$$

   Now group like terms. Write the x² terms, the x terms and the constant terms separately.

6. $$7x-3=0$$

   x² cancels, and simplifying gives 7x – 3 = 0. This is a linear equation, not a quadratic.

7. $$\text{Degree}=1\Rightarrow \text{NOT a Quadratic Equation}$$

   The highest power of x in this equation is 1, not 2. So this is not a quadratic equation.

Solution:

1. $$\text{Given: }(x+2{)}^3=2x(x^2-1)$$

   We have to check whether this equation is quadratic or not. First, expand the left side.

2. $$(x+2{)}^3=x^3+3(x^2)(2)+3(x)(4)+8=x^3+6x^2+12x+8$$

   On the left side, use the identity (a + b)³ = a³ + 3a²b + 3ab² + b³. Here a = x and b = 2.

3. $$\text{RHS: }2x(x^2-1)=2x^3-2x$$

   Now expand the right side: 2x(x² – 1) = 2x³ – 2x.

4. $$x^3+6x^2+12x+8=2x^3-2x$$

   Now set the LHS equal to the RHS and move all the terms to one side.

5. $$x^3+6x^2+12x+8-2x^3+2x=0$$

   Move all the right side terms to the left side. The signs will change.

6. $$-x^3+6x^2+14x+8=0⟹x^3-6x^2-14x-8=0$$

   Combining like terms gives -x³ + 6x² + 14x + 8 = 0. Multiplying by -1 gives x³ – 6x² – 14x – 8 = 0.

7. $$\text{Degree}=3\mathrm{\ne }2⟹\text{NOT a Quadratic Equation}$$

   The highest power of x in this equation is 3, so its degree is 3. A quadratic equation must have degree exactly 2, so this is not a quadratic equation.

Solution:

1. $$x^3-4x^2-x+1=(x-2{)}^3$$

   We have to check whether this equation is quadratic or not. First, expand the right side, (x – 2)³.

2. $$(x-2{)}^3=x^3-3\cdot x^2\cdot 2+3\cdot x\cdot 4-8$$

   The formula for (a – b)³ is a³ – 3a²b + 3ab² – b³. Here a = x and b = 2.

3. $$(x-2{)}^3=x^3-6x^2+12x-8$$

   After putting in the values, the right side becomes x³ – 6x² + 12x – 8.

4. $$x^3-4x^2-x+1=x^3-6x^2+12x-8$$

   Now write both sides. The next step is to move the right side to the left, so everything is on one side.

5. $$x^3-4x^2-x+1-x^3+6x^2-12x+8=0$$

   Move all the right side terms to the left side. The signs change.

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6. $$(1-1)x^3+(-4+6)x^2+(-1-12)x+(1+8)=0$$

   Now combine like terms. The x³ terms cancel, and we add the other terms.

7. $$2x^2-13x+9=0$$

   After simplifying we get 2x² – 13x + 9 = 0. It has an x² term and the highest power is 2.

8. $$\therefore \text{Degree}=2\Rightarrow \text{It IS a Quadratic Equation}$$

   The highest power in this equation is 2 and the coefficient of x² is 2, which is not zero. So this is a quadratic equation.

Solution:

1. $$\text{Rectangular plot: Area}=528{\text{ m}}^2$$

   In real life, when the area of a plot or room and a link between its sides are given, we form a quadratic equation exactly like this. Architects and civil engineers do the same.

2. $$\text{Let breadth}=x\text{ m, length}=(2x+1)\text{ m}$$

   First, choose the variable. Let the breadth be x metres. The length is one more than twice the breadth, so the length is 2x + 1 metres.

3. $$\text{Area of rectangle}=\text{length}\times \text{breadth}$$

   Remember the area formula of a rectangle: area = length × breadth.

4. $$(2x+1)\times x=528$$

   The given area is 528 square metres, so we form the equation (2x + 1) × x = 528.

5. $$2x^2+x=528$$

   Expand the left side. (2x + 1) × x gives 2x² + x.

6. $$2x^2+x-528=0$$

   Now move 528 to the left side. This gives the quadratic equation 2x² + x – 528 = 0.

Solution:

1. $$\text{Room dimensions: }x\times (x+1)=306\text{ sq units}$$

   Think of real life: if the length and width of a rectangular room are consecutive integers and its area is 306 square units, we get exactly this equation.

2. $$\text{Let 1st integer}=x,\text{2nd integer}=x+1$$

   Let the first positive integer be x. Then the next consecutive integer is x + 1.

3. $$x\times (x+1)=306$$

   The product of the two consecutive integers is 306, so we write x(x + 1) = 306.

4. $$x^2+x=306$$

   Expand the left side: x(x + 1) = x² + x.

5. $$x^2+x-306=0$$

   Now move 306 to the left side by subtracting 306 from both sides. This gives the quadratic equation.

6. $$\text{Here: }a=1,b=1,c=-306$$

   This is in the standard form ax² + bx + c = 0, where a = 1, b = 1 and c = -306.

Solution:

1. $$\text{Let Rohan’s present age}=x\text{ years}$$

   First, choose the variable. Let Rohan’s present age be x years.

2. $$\text{Mother’s present age}=x+26\text{ years}$$

   His mother is 26 years older than Rohan, so her present age is x + 26 years.

3. $$\text{Rohan’s age after 3 years}=x+3$$

   After 3 years, Rohan’s age will be x + 3, and his mother’s age will be x + 26 + 3, which is x + 29.

4. $$\text{Mother’s age after 3 years}=(x+26)+3=x+29$$

   So after 3 years, their ages are x + 3 and x + 29.

5. $$(x+3)(x+29)=360$$

   The question says the product of their ages after 3 years will be 360. So we write (x + 3)(x + 29) = 360.

6. $$x^2+29x+3x+87=360$$

   Now expand the left side. x(x + 29) + 3(x + 29) gives x² + 32x + 87 = 360.

7. $$x^2+32x+87-360=0$$

   Move 360 to the left side. 87 – 360 = -273.

8. $$x^2+32x-273=0$$

   This is our required quadratic equation: x² + 32x – 273 = 0.

Solution:

1. $$\text{Speed}\times \text{Time}=\text{Distance (Train Journey)}$$

   This is just like real life. When a train runs slower, it takes more time to cover the same distance. We are writing this idea as an equation here.

2. $$\text{Let speed of train}=x\text{ km/h}$$

   First, choose the variable. Let the speed of the train be x km per hour.

3. $$\text{Time}=\frac{\text{Distance}}{\text{Speed}}=\frac{480}{x}\text{ hours}$$

   We know that Time = Distance ÷ Speed, so the time at normal speed is 480/x hours.

4. $$\text{Reduced speed}=(x-8)\text{ km/h},\text{New time}=\frac{480}{x-8}\text{ hours}$$

   If the speed drops by 8 km per hour, the new speed is x – 8, and the time at this speed is 480/(x – 8) hours.

5. $$\frac{480}{x-8}-\frac{480}{x}=3$$

   The question says the new time is 3 hours more than the old time. This gives the equation.

6. $$480x-480(x-8)=3\cdot x(x-8)$$

   Multiply both sides by x(x – 8), and open the brackets.

7. $$480x-480x+3840=3x^2-24x$$

   On the left side, 480x – 480x cancels and only 3840 is left. The right side gives 3x² – 24x.

8. $$3x^2-24x-3840=0⟹x^2-8x-1280=0$$

   Move all the terms to the left side, then divide the whole equation by 3. The final quadratic equation is x² – 8x – 1280 = 0.