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:
$$x^2 + 2x + 1 = 2x - 6$$$$x^2 + 2x + 1 - 2x + 6 = 0$$$$x^2 + 7 = 0$$This is of the form $ax^2 + bx + c = 0$ with $a = 1, b = 0, c = 7$.
Since $a \neq 0$ and the highest degree is $2$, this is a quadratic equation.
(ii) $x^2 - 2x = (-2)(3-x)$
Expanding the right side:
$$x^2 - 2x = -6 + 2x$$$$x^2 - 2x - 2x + 6 = 0$$$$x^2 - 4x + 6 = 0$$This is of the form $ax^2 + bx + c = 0$ with $a = 1$.
This is a quadratic equation.
(iii) $(x-2)(x+1) = (x-1)(x+3)$
Expanding the left side: $x^2 + x - 2x - 2 = x^2 - x - 2$
Expanding the right side: $x^2 + 3x - x - 3 = x^2 + 2x - 3$
Setting LHS $=$ RHS:
$$x^2 - x - 2 = x^2 + 2x - 3$$$$-x - 2 - 2x + 3 = 0$$$$-3x + 1 = 0$$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 LHS: $2x^2 + x - 6x - 3 = 2x^2 - 5x - 3$
Expanding RHS: $x^2 + 5x$
Setting equal:
$$2x^2 - 5x - 3 = x^2 + 5x$$$$x^2 - 10x - 3 = 0$$This is a quadratic equation.
(v) $(2x-1)(x-3) = (x+5)(x-1)$
Expanding LHS: $2x^2 - 6x - x + 3 = 2x^2 - 7x + 3$
Expanding RHS: $x^2 - x + 5x - 5 = x^2 + 4x - 5$
Setting equal:
$$2x^2 - 7x + 3 = x^2 + 4x - 5$$$$x^2 - 11x + 8 = 0$$This is a quadratic equation.
(vi) $x^2 + 3x + 1 = (x-2)^2$
Expanding the right side:
$$x^2 + 3x + 1 = x^2 - 4x + 4$$$$x^2 + 3x + 1 - x^2 + 4x - 4 = 0$$$$7x - 3 = 0$$The $x^2$ terms cancel.
This is NOT a quadratic equation (it is linear).
(vii) $(x+2)^3 = 2x(x^2 - 1)$
Expanding LHS: $(x+2)^3 = x^3 + 6x^2 + 12x + 8$
Expanding RHS: $2x^3 - 2x$
Setting equal:
$$x^3 + 6x^2 + 12x + 8 = 2x^3 - 2x$$$$x^3 - 6x^2 - 14x - 8 = 0$$The highest degree is $3$ (cubic).
This is NOT a quadratic equation (it is cubic).
(viii) $x^3 - 4x^2 - x + 1 = (x-2)^3$
Expanding RHS: $(x-2)^3 = x^3 - 6x^2 + 12x - 8$
Setting LHS $=$ RHS:
$$x^3 - 4x^2 - x + 1 = x^3 - 6x^2 + 12x - 8$$$$-4x^2 + 6x^2 - x - 12x + 1 + 8 = 0$$$$2x^2 - 13x + 9 = 0$$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 = $x$ metres.
Then the length = $(2x + 1)$ metres.
Area = length × breadth:
$$x(2x + 1) = 528$$$$2x^2 + x = 528$$$$2x^2 + x - 528 = 0$$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 = $x$.
Then the next consecutive integer = $x + 1$.
$$x(x+1) = 306$$$$x^2 + x = 306$$$$x^2 + x - 306 = 0$$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 = $x$ years.
Mother’s present age = $(x + 26)$ years.
Three years from now:
- Rohan’s age = $x + 3$
- Mother’s age = $x + 29$
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.
Let the speed of the train = $x$ km/h.
Time taken at speed $x$ = $\dfrac{480}{x}$ hours.
Time taken at speed $(x-8)$ = $\dfrac{480}{x-8}$ hours.
Given that the slower speed takes 3 more hours:
$$\frac{480}{x-8} - \frac{480}{x} = 3$$$$480\left(\frac{x - (x-8)}{x(x-8)}\right) = 3$$$$480 \times \frac{8}{x(x-8)} = 3$$$$\frac{3840}{x(x-8)} = 3$$$$x(x-8) = 1280$$$$x^2 - 8x - 1280 = 0$$This is the required quadratic equation.