NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1#
Welcome to our NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1. This introductory exercise helps students identify quadratic equations and represent real-life situations in the form of quadratic equations. A thorough understanding of this exercise forms the foundation for all subsequent methods of solving quadratic equations.
Quadratic Equation: An equation of the form $ax^2 + bx + c = 0$, where $a \neq 0$ and $a$, $b$, $c$ are real numbers, is called a quadratic equation. The degree of the equation must be exactly 2. If the degree is 1 or 0, it is not quadratic.
1. Check whether the following are quadratic equations: #
(i) $(x+1)^2 = 2(x-3)$
Expanding and bringing all terms to the left side:
$$(x+1)^2 = 2(x-3)$$$$\Rightarrow x^2 + 2x + 1 = 2x - 6$$
$$\Rightarrow x^2 + 2x + 1 - 2x + 6 = 0$$
$$\Rightarrow x^2 + 7 = 0$$
This is of the form $ax^2 + bx + c = 0$ with $a = 1,\ b = 0,\ c = 7$.
∴ Since $a \neq 0$ and the highest power is $2$, this is a quadratic equation.
(ii) $x^2 - 2x = (-2)(3-x)$
Expanding the right side and bringing all terms to the left side:
$$x^2 - 2x = (-2)(3-x)$$$$\Rightarrow x^2 - 2x = -6 + 2x$$
$$\Rightarrow x^2 - 2x - 2x + 6 = 0$$
$$\Rightarrow x^2 - 4x + 6 = 0$$
This is of the form $ax^2 + bx + c = 0$ with $a = 1,\ b = -4,\ c = 6$.
∴ Since $a \neq 0$ and the highest power is $2$, this is a quadratic equation.
(iii) $(x-2)(x+1) = (x-1)(x+3)$
Expanding the left side $(x-2)(x+1)$:
$$= x^2 + x - 2x - 2$$$$= x^2 - x - 2$$
Expanding the right side $(x-1)(x+3)$:
$$= x^2 + 3x - x - 3$$$$= x^2 + 2x - 3$$
Setting LHS $=$ RHS and bringing all terms to the left side:
$$x^2 - x - 2 = x^2 + 2x - 3$$$$\Rightarrow x^2 - x - 2 - x^2 - 2x + 3 = 0$$
$$\Rightarrow -3x + 1 = 0$$
The $x^2$ terms cancel, leaving a linear equation.
∴ This is NOT a quadratic equation (it is a linear equation).
(iv) $(x-3)(2x+1) = x(x+5)$
Expanding the left side $(x-3)(2x+1)$:
$$= 2x^2 + x - 6x - 3$$$$= 2x^2 - 5x - 3$$
Expanding the right side $x(x+5)$:
$$= x^2 + 5x$$Setting LHS $=$ RHS and bringing all terms to the left side:
$$2x^2 - 5x - 3 = x^2 + 5x$$$$\Rightarrow 2x^2 - 5x - 3 - x^2 - 5x = 0$$
$$\Rightarrow x^2 - 10x - 3 = 0$$
∴ This is a quadratic equation.
(v) $(2x-1)(x-3) = (x+5)(x-1)$
Expanding the left side $(2x-1)(x-3)$:
$$= 2x^2 - 6x - x + 3$$$$= 2x^2 - 7x + 3$$
Expanding the right side $(x+5)(x-1)$:
$$= x^2 - x + 5x - 5$$$$= x^2 + 4x - 5$$
Setting LHS $=$ RHS and bringing all terms to the left side:
$$2x^2 - 7x + 3 = x^2 + 4x - 5$$$$\Rightarrow 2x^2 - 7x + 3 - x^2 - 4x + 5 = 0$$
$$\Rightarrow x^2 - 11x + 8 = 0$$
∴ This is a quadratic equation.
(vi) $x^2 + 3x + 1 = (x-2)^2$
Expanding the right side $(x-2)^2$:
$$= x^2 - 4x + 4$$Setting LHS $=$ RHS and bringing all terms to the left side:
$$x^2 + 3x + 1 = x^2 - 4x + 4$$$$\Rightarrow x^2 + 3x + 1 - x^2 + 4x - 4 = 0$$
$$\Rightarrow 7x - 3 = 0$$
The $x^2$ terms cancel, leaving a linear equation.
∴ This is NOT a quadratic equation (it is linear).
(vii) $(x+2)^3 = 2x(x^2 - 1)$
Expanding the left side $(x+2)^3$:
$$= x^3 + 6x^2 + 12x + 8$$Expanding the right side $2x(x^2 - 1)$:
$$= 2x^3 - 2x$$Setting LHS $=$ RHS and bringing all terms to the right side:
$$x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$$$$\Rightarrow 0 = 2x^3 - 2x - x^3 - 6x^2 - 12x - 8$$
$$\Rightarrow x^3 - 6x^2 - 14x - 8 = 0$$
The highest power is $3$ (cubic).
∴ This is NOT a quadratic equation (it is cubic).
(viii) $x^3 - 4x^2 - x + 1 = (x-2)^3$
Expanding the right side $(x-2)^3$:
$$= x^3 - 6x^2 + 12x - 8$$Setting LHS $=$ RHS and bringing all terms to the left side:
$$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$$$$\Rightarrow x^3 - 4x^2 - x + 1 - x^3 + 6x^2 - 12x + 8 = 0$$
$$\Rightarrow 2x^2 - 13x + 9 = 0$$
∴ This is a quadratic equation.
2. Represent the following situations in the form of quadratic equations: #
(i) The area of a rectangular plot is 528 m². The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Let the breadth of the plot be $x$ metres.
The length is one more than twice the breadth, so:
$$\text{length} = (2x + 1) \text{ metres}$$Area = length × breadth, and the area is 528 m², so:
$$x(2x + 1) = 528$$$$\Rightarrow 2x^2 + x = 528$$
$$\Rightarrow 2x^2 + x - 528 = 0$$
∴ This is the required quadratic equation.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Let the first positive integer be $x$.
The next consecutive integer is $x + 1$.
Their product is 306, so:
$$x(x+1) = 306$$$$\Rightarrow x^2 + x = 306$$
$$\Rightarrow x^2 + x - 306 = 0$$
∴ This is the required quadratic equation.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
Let Rohan’s present age be $x$ years.
His mother is 26 years older, so her present age is $(x + 26)$ years.
3 years from now:
- Rohan’s age will be $(x + 3)$ years.
- His mother’s age will be $(x + 29)$ years.
The product of their ages will be 360, so:
$$(x+3)(x+29) = 360$$$$\Rightarrow x^2 + 29x + 3x + 87 = 360$$
$$\Rightarrow x^2 + 32x + 87 - 360 = 0$$
$$\Rightarrow x^2 + 32x - 273 = 0$$
∴ This is the required quadratic equation.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Let the speed of the train be $x$ km/h.
Time = distance ÷ speed.
At speed $x$ km/h, the time taken is:
$$\dfrac{480}{x} \text{ hours}$$At speed $(x-8)$ km/h, the time taken is:
$$\dfrac{480}{x-8} \text{ hours}$$The slower speed takes 3 hours more, so:
$$\dfrac{480}{x-8} - \dfrac{480}{x} = 3$$$$\Rightarrow 480\left(\dfrac{x - (x-8)}{x(x-8)}\right) = 3$$
$$\Rightarrow 480 \times \dfrac{8}{x(x-8)} = 3$$
$$\Rightarrow \dfrac{3840}{x(x-8)} = 3$$
$$\Rightarrow x(x-8) = 1280$$
$$\Rightarrow x^2 - 8x - 1280 = 0$$
∴ This is the required quadratic equation.