CBSE Class 10 Mathematics Chapter 4: Quadratic Equations NCERT Solutions

A quadratic equation is an equation of the form $ax^2\; +\; bx\; +\; c\; =\; 0$, where $a$, $b$, and $c$ are real numbers and $a \neq 0$. We need to simplify each given equation to check if it fits this form.

1. Given equation: $(x+1)^2\; =\; 2(x-3)$

   Expand both sides: $x^2\; +\; 2x\; +\; 1\; =\; 2x\; -\; 6$

   Rearrange terms to one side: $x^2\; +\; 2x\; +\; 1\; -\; 2x\; +\; 6\; =\; 0$

   Simplify: $x^2\; +\; 7\; =\; 0$

   This equation is in the form $ax^2\; +\; bx\; +\; c\; =\; 0$ with $a=1$, $b=0$, and $c=7$. Since $a \neq 0$, it is a quadratic equation.

2. Given equation: $x^2\; -\; 2x\; =\; (-2)(3-x)$

   Expand the right side: $x^2\; -\; 2x\; =\; -6\; +\; 2x$

   Rearrange terms: $x^2\; -\; 2x\; -\; 2x\; +\; 6\; =\; 0$

   Simplify: $x^2\; -\; 4x\; +\; 6\; =\; 0$

   This equation is in the form $ax^2\; +\; bx\; +\; c\; =\; 0$ with $a=1$, $b=-4$, and $c=6$. Since $a \neq 0$, it is a quadratic equation.

3. Given equation: $(x-2)(x+1)\; =\; (x-1)(x+3)$

   Expand both sides: $x^2\; +\; x\; -\; 2x\; -\; 2\; =\; x^2\; +\; 3x\; -\; x\; -\; 3$

   Simplify each side: $x^2\; -\; x\; -\; 2\; =\; x^2\; +\; 2x\; -\; 3$

   Rearrange terms: $x^2\; -\; x\; -\; 2\; -\; x^2\; -\; 2x\; +\; 3\; =\; 0$

   Simplify: $-3x\; +\; 1\; =\; 0$ or $3x\; -\; 1\; =\; 0$

   The highest power of $x$ is 1. This equation is linear, not quadratic. Therefore, it is not a quadratic equation.

4. Given equation: $(x-3)(2x+1)\; =\; x(x+5)$

   Expand both sides: $2x^2\; +\; x\; -\; 6x\; -\; 3\; =\; x^2\; +\; 5x$

   Simplify the left side: $2x^2\; -\; 5x\; -\; 3\; =\; x^2\; +\; 5x$

   Rearrange terms: $2x^2\; -\; 5x\; -\; 3\; -\; x^2\; -\; 5x\; =\; 0$

   Simplify: $x^2\; -\; 10x\; -\; 3\; =\; 0$

   This equation is in the form $ax^2\; +\; bx\; +\; c\; =\; 0$ with $a=1$, $b=-10$, and $c=-3$. Since $a \neq 0$, it is a quadratic equation.

5. Given equation: $(2x-1)(x-3)\; =\; (x+5)(x-1)$

   Expand both sides: $2x^2\; -\; 6x\; -\; x\; +\; 3\; =\; x^2\; -\; x\; +\; 5x\; -\; 5$

   Simplify each side: $2x^2\; -\; 7x\; +\; 3\; =\; x^2\; +\; 4x\; -\; 5$

   Rearrange terms: $2x^2\; -\; 7x\; +\; 3\; -\; x^2\; -\; 4x\; +\; 5\; =\; 0$

   Simplify: $x^2\; -\; 11x\; +\; 8\; =\; 0$

   This equation is in the form $ax^2\; +\; bx\; +\; c\; =\; 0$ with $a=1$, $b=-11$, and $c=8$. Since $a \neq 0$, it is a quadratic equation.

6. Given equation: $x^2\; +\; 3x\; +\; 1\; =\; (x-2)^2$

   Expand the right side: $x^2\; +\; 3x\; +\; 1\; =\; x^2\; -\; 4x\; +\; 4$

   Rearrange terms: $x^2\; +\; 3x\; +\; 1\; -\; x^2\; +\; 4x\; -\; 4\; =\; 0$

   Simplify: $7x\; -\; 3\; =\; 0$

   The highest power of $x$ is 1. This equation is linear, not quadratic. Therefore, it is not a quadratic equation.

7. Given equation: $(x+2)^3\; =\; 2x(x^2-1)$

   Expand the left side using $(a+b)^3\; =\; a^3\; +\; b^3\; +\; 3a^2b\; +\; 3ab^2$: $x^3\; +\; 2^3\; +\; 3(x^2)(2)\; +\; 3(x)(2^2)\; =\; 2x^3\; -\; 2x$

   $x^3\; +\; 8\; +\; 6x^2\; +\; 12x\; =\; 2x^3\; -\; 2x$

   Rearrange terms: $x^3\; +\; 6x^2\; +\; 12x\; +\; 8\; -\; 2x^3\; +\; 2x\; =\; 0$

   Simplify: $-x^3\; +\; 6x^2\; +\; 14x\; +\; 8\; =\; 0$

   The highest power of $x$ is 3. This is a cubic equation, not a quadratic equation. Therefore, it is not a quadratic equation.

8. Given equation: $x^3\; -\; 4x^2\; -\; x\; +\; 1\; =\; (x-2)^3$

   Expand the right side using $(a-b)^3\; =\; a^3\; -\; b^3\; -\; 3a^2b\; +\; 3ab^2$: $x^3\; -\; 4x^2\; -\; x\; +\; 1\; =\; x^3\; -\; 2^3\; -\; 3(x^2)(2)\; +\; 3(x)(2^2)$

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

   Rearrange terms: $x^3\; -\; 4x^2\; -\; x\; +\; 1\; -\; x^3\; +\; 8\; +\; 6x^2\; -\; 12x\; =\; 0$

   Simplify: $(-4x^2\; +\; 6x^2)\; +\; (-x\; -\; 12x)\; +\; (1\; +\; 8)\; =\; 0$

   $2x^2\; -\; 13x\; +\; 9\; =\; 0$

   This equation is in the form $ax^2\; +\; bx\; +\; c\; =\; 0$ with $a=2$, $b=-13$, and $c=9$. Since $a \neq 0$, it is a quadratic equation.

1. Let the breadth of the rectangular plot be $x$ metres.

   According to the problem, the length of the plot is one more than twice its breadth. So, the length is $(2x\; +\; 1)$ metres.

   The area of a rectangle is given by the formula: Area = Length $\backslash times$ Breadth.

   We are given that the area is 528 m². Therefore,

   $528\; =\; (2x\; +\; 1)\; \backslash times\; x$

   Expand the equation: $528\; =\; 2x^2\; +\; x$

   Rearrange the terms to form a quadratic equation in standard form ($ax^2\; +\; bx\; +\; c\; =\; 0$):

   $2x^2\; +\; x\; -\; 528\; =\; 0$

   This is the required quadratic equation representing the situation.

2. Let the first positive integer be $x$.

   Since the integers are consecutive positive integers, the next integer will be $x\; +\; 1$.

   The problem states that the product of these two integers is 306.

   So, we can write the equation as:

   $x(x\; +\; 1)\; =\; 306$

   Expand the equation:

   $x^2\; +\; x\; =\; 306$

   Rearrange the terms to form a quadratic equation in standard form:

   $x^2\; +\; x\; -\; 306\; =\; 0$

   This is the required quadratic equation representing the situation.

3. Let Rohan's present age be $x$ years.

   Rohan's mother is 26 years older than him, so her present age is $(x\; +\; 26)$ years.

   In 3 years from now:

   Rohan's age will be $(x\; +\; 3)$ years.

   Rohan's mother's age will be $(x\; +\; 26\; +\; 3)\; =\; (x\; +\; 29)$ years.

   The problem states that the product of their ages 3 years from now will be 360.

   So, we can write the equation as:

   $(x\; +\; 3)(x\; +\; 29)\; =\; 360$

   Expand the equation:

   $x^2\; +\; 29x\; +\; 3x\; +\; 87\; =\; 360$

   $x^2\; +\; 32x\; +\; 87\; =\; 360$

   Rearrange the terms to form a quadratic equation in standard form:

   $x^2\; +\; 32x\; +\; 87\; -\; 360\; =\; 0$

   $x^2\; +\; 32x\; -\; 273\; =\; 0$

   This is the required quadratic equation representing the situation.

4. Let the uniform speed of the train be $x$ km/h.

   The distance to be traveled is 480 km.

   The time taken to cover the distance at uniform speed is given by Time = Distance / Speed.

   So, the original time taken is $\backslash frac\{480\}\{x\}$ hours.

   According to the problem, if the speed had been 8 km/h less, the new speed would be $(x\; -\; 8)$ km/h.

   The time taken to cover the same distance at this reduced speed would be $\backslash frac\{480\}\{x-8\}$ hours.

   The problem states that this new time is 3 hours more than the original time.

   Therefore, we can write the equation as:

   $\backslash frac\{480\}\{x-8\}\; =\; \backslash frac\{480\}\{x\}\; +\; 3$

   To solve this, first, let's get a common denominator on the right side:

   $\backslash frac\{480\}\{x-8\}\; =\; \backslash frac\{480\; +\; 3x\}\{x\}$

   Now, cross-multiply:

   $480x\; =\; (x-8)(480\; +\; 3x)$

   Expand the right side:

   $480x\; =\; 480x\; +\; 3x^2\; -\; 8(480)\; -\; 8(3x)$

   $480x\; =\; 480x\; +\; 3x^2\; -\; 3840\; -\; 24x$

   Rearrange all terms to one side to form a quadratic equation:

   $480x\; -\; 480x\; -\; 3x^2\; +\; 3840\; +\; 24x\; =\; 0$

   $-3x^2\; +\; 24x\; +\; 3840\; =\; 0$

   Divide the entire equation by -3 to simplify:

   $x^2\; -\; 8x\; -\; 1280\; =\; 0$

   This is the required quadratic equation representing the situation.