Get detailed, step-by-step solutions for NCERT Class 10 Maths Chapter 4 Exercise 4.1 on Quadratic Equations. Class 10 Maths Exercise 4.1, Check Quadratic Equation, Form Quadratic Equations, Standard Form $ax^2+bx+c=0$, Word Problems Quadratic Equations, Speed Time Distance Problem, Rectangular Plot Area
Learn to check if an equation is quadratic by simplifying it to the standard form $\mathbf{ax^2 + bx + c = 0}$ where $a \neq 0$ (Q.1). Solutions cover tricky expansions like $(x+2)^3$ and $(x-2)^2$. Master forming quadratic equations from real-life scenarios, including area problems, consecutive integers, age problems, and speed-time-distance problems (Q.2). Essential for setting up quadratic models.
This exercise covers checking whether given equations are quadratic and formulating quadratic equations from word problems.
1. Check whether the following are quadratic equations
A quadratic equation is any equation that can be rearranged into the standard form $\mathbf{ax^2 + bx + c = 0}$, where $x$ is the variable, and $a, b, c$ are real numbers with $\mathbf{a \neq 0}$. The key characteristic is that the highest power of the variable is 2.
(i) $(x + 1)^2 = 2(x – 3)$
$$x^2 + 2x + 1 = 2x – 6$$
$$x^2 + 2x – 2x + 1 + 6 = 0$$
$$\mathbf{x^2 + 7 = 0}$$
This is in the form $ax^2 + bx + c = 0$ with $a=1, b=0, c=7$.
Yes, it is a quadratic equation.
(ii) $x^2 – 2x = (–2) (3 – x)$
$$x^2 – 2x = -6 + 2x$$
$$x^2 – 2x – 2x + 6 = 0$$
$$\mathbf{x^2 – 4x + 6 = 0}$$
This is in the form $ax^2 + bx + c = 0$ with $a=1, b=-4, c=6$.
Yes, it is a quadratic equation.
(iii) $(x – 2)(x + 1) = (x – 1)(x + 3)$
$$x^2 + x – 2x – 2 = x^2 + 3x – x – 3$$
$$x^2 – x – 2 = x^2 + 2x – 3$$
Subtract $x^2$ from both sides:
$$-x – 2 = 2x – 3$$
$$-3x + 1 = 0$$
The highest power of $x$ is 1 ($a=0$).
No, it is not a quadratic equation (it is a linear equation).
(iv) $(x – 3)(2x + 1) = x(x + 5)$
$$2x^2 + x – 6x – 3 = x^2 + 5x$$
$$2x^2 – 5x – 3 = x^2 + 5x$$
$$2x^2 – x^2 – 5x – 5x – 3 = 0$$
$$\mathbf{x^2 – 10x – 3 = 0}$$
This is in the form $ax^2 + bx + c = 0$ with $a=1$.
Yes, it is a quadratic equation.
(v) $(2x – 1)(x – 3) = (x + 5)(x – 1)$
$$2x^2 – 6x – x + 3 = x^2 – x + 5x – 5$$
$$2x^2 – 7x + 3 = x^2 + 4x – 5$$
$$2x^2 – x^2 – 7x – 4x + 3 + 5 = 0$$
$$\mathbf{x^2 – 11x + 8 = 0}$$
This is in the form $ax^2 + bx + c = 0$ with $a=1$.
Yes, it is a quadratic equation.
(vi) $x^2 + 3x + 1 = (x – 2)^2$
$$x^2 + 3x + 1 = x^2 – 4x + 4$$
Subtract $x^2$ from both sides:
$$3x + 1 = -4x + 4$$
$$7x – 3 = 0$$
The highest power of $x$ is 1 ($a=0$).
No, it is not a quadratic equation (it is a linear equation).
(vii) $(x + 2)^3 = 2x (x^2 – 1)$
Expand the left side using the identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$:
$$x^3 + 3(x)^2(2) + 3(x)(2)^2 + 2^3 = 2x^3 – 2x$$
$$x^3 + 6x^2 + 12x + 8 = 2x^3 – 2x$$
$$0 = 2x^3 – x^3 – 6x^2 – 2x – 12x – 8$$
$$\mathbf{x^3 – 6x^2 – 14x – 8 = 0}$$
The highest power of $x$ is 3.
No, it is not a quadratic equation (it is a cubic equation).
(viii) $x^3 – 4x^2 – x + 1 = (x – 2)^3$
Expand the right side using the identity $(a-b)^3 = a^3 – 3a^2b + 3ab^2 – b^3$:
$$x^3 – 4x^2 – x + 1 = x^3 – 3(x)^2(2) + 3(x)(2)^2 – 2^3$$
$$x^3 – 4x^2 – x + 1 = x^3 – 6x^2 + 12x – 8$$
Subtract $x^3$ from both sides:
$$-4x^2 – x + 1 = -6x^2 + 12x – 8$$
$$-4x^2 + 6x^2 – x – 12x + 1 + 8 = 0$$
$$\mathbf{2x^2 – 13x + 9 = 0}$$
This is in the form $ax^2 + bx + c = 0$ with $a=2$.
Yes, it is a quadratic equation.
2. Represent the following situations in the form of quadratic equations
(i) Rectangular Plot Area
Let the breadth of the plot be $x$ meters.
The length is one more than twice the breadth: $2x + 1$ meters.
Area = Length $\times$ Breadth
$$x(2x + 1) = 528$$
$$2x^2 + x = 528$$
$$\mathbf{2x^2 + x – 528 = 0}$$
(ii) Product of Two Consecutive Positive Integers
Let the first positive integer be $x$.
The next consecutive positive integer is $x + 1$.
The product is 306:
$$x(x + 1) = 306$$
$$x^2 + x = 306$$
$$\mathbf{x^2 + x – 306 = 0}$$
(iii) Rohan’s Age
Let Rohan’s present age be $x$ years.
Mother’s present age: $x + 26$ years.
Ages 3 years from now:
- Rohan’s age: $x + 3$
- Mother’s age: $(x + 26) + 3 = x + 29$The product of their ages 3 years from now will be 360:$$(x + 3)(x + 29) = 360$$$$x^2 + 29x + 3x + 87 = 360$$$$x^2 + 32x + 87 – 360 = 0$$$$\mathbf{x^2 + 32x – 273 = 0}$$
(iv) Train Speed and Time
Let the uniform speed of the train be $x$ km/h.
Distance traveled is 480 km.
Time taken = $\text{Distance} / \text{Speed}$.
- Original Time ($T_1$): $T_1 = \frac{480}{x}$ hours.
- New Speed: $x – 8$ km/h.
- New Time ($T_2$): $T_2 = \frac{480}{x – 8}$ hours.
The new time is 3 hours more than the original time: $T_2 = T_1 + 3$.
$$\frac{480}{x – 8} = \frac{480}{x} + 3$$
To form the quadratic equation, rearrange the terms:
$$\frac{480}{x – 8} – \frac{480}{x} = 3$$
Multiply by the common denominator $x(x – 8)$:
$$480x – 480(x – 8) = 3x(x – 8)$$
$$480x – 480x + 3840 = 3x^2 – 24x$$
$$0 = 3x^2 – 24x – 3840$$
Divide the entire equation by 3 to simplify:
$$\mathbf{x^2 – 8x – 1280 = 0}$$