This page provides comprehensive Quadratic Equations – Exercise 4.1. Free NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations Exercise 4-1. Step-by-step explained answers for CBSE Board exams. Download PDF and practice now.
NCERT Exercise Practice questions with detailed step-by-step solutions
Check whether the following are quadratic equations:
(i) $(x + 1)^2 = 2(x - 3)$
(ii) $x^2 - 2x = (-2)(3 - x)$
(iii) $(x - 2)(x + 1) = (x - 1)(x + 3)$
(iv) $(x - 3)(2x + 1) = x(x + 5)$
(v) $(2x - 1)(x - 3) = (x + 5)(x - 1)$
(vi) $x^2 + 3x + 1 = (x - 2)^2$
(vii) $(x + 2)^3 = 2x(x^2 - 1)$
(viii) $x^3 - 4x^2 - x + 1 = (x - 2)^3$
(i)
$(x + 1)^2 = 2(x - 3) \Rightarrow x^2 + 2x + 1 = 2x - 6$
$x^2 + 2x - 2x + 1 + 6 = 0 \Rightarrow x^2 + 7 = 0$.
It is of the form $ax^2 + bx + c = 0$.
✅ Yes, it is a quadratic equation.(ii)
$x^2 - 2x = (-2)(3 - x) \Rightarrow x^2 - 2x = -6 + 2x$
$x^2 - 2x - 2x + 6 = 0 \Rightarrow x^2 - 4x + 6 = 0$.
✅ Yes, it is a quadratic equation.(iii)
$(x - 2)(x + 1) = (x - 1)(x + 3)$
$x^2 + x - 2x - 2 = x^2 + 3x - x - 3$
$x^2 - x - 2 = x^2 + 2x - 3$
$-x - 2x - 2 + 3 = 0 \Rightarrow -3x + 1 = 0$.
No $x^2$ term.
✅ No, it is not a quadratic equation.(iv)
$(x - 3)(2x + 1) = x(x + 5)$
$2x^2 + x - 6x - 3 = x^2 + 5x$
$2x^2 - 5x - 3 - x^2 - 5x = 0 \Rightarrow x^2 - 10x - 3 = 0$.
✅ Yes, it is a quadratic equation.(v)
$(2x - 1)(x - 3) = (x + 5)(x - 1)$
$2x^2 - 6x - x + 3 = x^2 - x + 5x - 5$
$2x^2 - 7x + 3 = x^2 + 4x - 5$
$x^2 - 11x + 8 = 0$.
✅ Yes, it is a quadratic equation.(vi)
$x^2 + 3x + 1 = (x - 2)^2 = x^2 - 4x + 4$
$3x + 4x + 1 - 4 = 0 \Rightarrow 7x - 3 = 0$.
No $x^2$ term.
✅ No, it is not a quadratic equation.(vii)
$(x + 2)^3 = 2x(x^2 - 1)$
$x^3 + 8 + 6x(x + 2) = 2x^3 - 2x$
$x^3 + 8 + 6x^2 + 12x = 2x^3 - 2x$
$-x^3 + 6x^2 + 14x + 8 = 0$.
Degree is 3 (Cubic).
✅ No, it is not a quadratic equation.(viii)
$x^3 - 4x^2 - x + 1 = (x - 2)^3$
$x^3 - 4x^2 - x + 1 = x^3 - 8 - 6x(x - 2)$
$x^3 - 4x^2 - x + 1 = x^3 - 8 - 6x^2 + 12x$
$-4x^2 + 6x^2 - x - 12x + 1 + 8 = 0$
$2x^2 - 13x + 9 = 0$.
✅ Yes, it is a quadratic equation.
Question 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.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(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.
(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.
(i) Let breadth = $x$ m.
Length = $2x + 1$ m.
Area = Length $\times$ Breadth = $x(2x + 1) = 528$.
$2x^2 + x - 528 = 0$.
✅ Equation: $2x^2 + x - 528 = 0$(ii) Let the consecutive integers be $x$ and $x + 1$.
Product = $x(x + 1) = 306$.
$x^2 + x - 306 = 0$.
✅ Equation: $x^2 + x - 306 = 0$(iii) Let Rohan's age = $x$. Mother's age = $x + 26$.
After 3 years:
Rohan = $x + 3$, Mother = $(x + 26) + 3 = x + 29$.
Product = $(x + 3)(x + 29) = 360$.
$x^2 + 29x + 3x + 87 = 360$
$x^2 + 32x + 87 - 360 = 0 \Rightarrow x^2 + 32x - 273 = 0$.
✅ Equation: $x^2 + 32x - 273 = 0$(iv) Let speed of train = $x$ km/h.
Distance = 480 km. Time taken = $\frac{480}{x}$.
New speed = $x - 8$ km/h. New time = $\frac{480}{x - 8}$.
Difference in time = 3 hours.
$\frac{480}{x - 8} - \frac{480}{x} = 3$
$480 \left( \frac{x - (x - 8)}{x(x - 8)} \right) = 3$
$480(8) = 3x(x - 8)$
$3840 = 3x^2 - 24x$
Divide by 3: $1280 = x^2 - 8x \Rightarrow x^2 - 8x - 1280 = 0$.
✅ Equation: $x^2 - 8x - 1280 = 0$