NCERT Solutions For Class 10 Maths Chapter 4 Exercise 4.1 - Quadratic Equations

Class 10 Maths Chapter 4 Exercise 4.1

1. Check whether the following are quadratic equations:

(i) ${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

Ans: We are given an equation: ${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

We will simplify the given equation.

${{\left( \text{x + 1} \right)}^{\text{2}}}\text{ = 2}\left( \text{x - 3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 2x + 1 = 2x - 6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 1 = - 6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 7 = 0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

(ii) ${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

Ans: We are given an equation: ${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

We will simplify the given equation.

${{\text{x}}^{\text{2}}}\text{ - 2x = }\left( \text{-2} \right)\left( \text{3 - x} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 2x = - 6 + 2x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 4x + 6 =0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

(iii) $\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

Ans: We are given an equation: $\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

We will simplify the given equation.

$\left( \text{x - 2} \right)\left( \text{x + 1} \right)\text{ = }\left( \text{x - 1} \right)\left( \text{x + 3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 2x + x - 2 = }{{\text{x}}^{\text{2}}}\text{ - x + 3x - 3}$

$\Rightarrow \text{- 2x + x - 2 = - x + 3x - 3}$

$\Rightarrow \text{- x - 2 = 2x - 3}$

$\Rightarrow \text{3x - 1 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.

(iv) $\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

Ans: We are given an equation: $\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

We will simplify the given equation.

$\left( \text{x - 3} \right)\left( \text{2x + 1} \right)\text{ = x}\left( \text{x + 5} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{- 6x + x - 3 = }{{\text{x}}^{\text{2}}}\text{ + 5x}$

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{- 10x - 3 = 0}\]

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

(v) $\left( \text{2x - 1} \right)\left( \text{x - 3} \right)\text{ = }\left( \text{x + 5} \right)\left( \text{x - 1} \right)$

Ans: We are given an equation: $\left( \text{2x - 1} \right)\left( \text{x - 3} \right)\text{ = }\left( \text{x + 5} \right)\left( \text{x - 1} \right)$

We will simplify the given equation.

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ - x - 6x + 3 = }{{\text{x}}^{\text{2}}}\text{ + 5x - x - 5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 7x + 3 = 4x - 5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 11x + 8 = 0}$

The equation obtained after simplifying is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

(vi) ${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

Ans: We are given an equation: ${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

We will simplify the given equation.

${{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\left( \text{x - 2} \right)}^{\text{2}}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 3x + 1 = }{{\text{x}}^{\text{2}}}\text{ - 4x + 4}$

$\Rightarrow \text{3x + 1 = - 4x + 4}$

$\Rightarrow \text{7x - 3 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.

(vii) ${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

Ans: We are given an equation: ${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

We will simplify the given equation.

${{\left( \text{x + 2} \right)}^{\text{3}}}\text{ = 2x}\left( {{\text{x}}^{\text{2}}}\text{ - 1} \right)$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ + 6}{{\text{x}}^{\text{2}}}\text{ + 12x + 8 = 2}{{\text{x}}^{\text{3}}}\text{ - 2x}$

$\Rightarrow \text{6}{{\text{x}}^{\text{2}}}\text{ + 14x + 8 = }{{\text{x}}^{\text{3}}}$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ - 6}{{\text{x}}^{\text{2}}}\text{ - 14x - 8 = 0}$

This equation obtained is not of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is not a quadratic equation.

(viii) \[{{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\left( \text{x - 2} \right)}^{\text{3}}}\]

Ans: We are given an equation: ${{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\left( \text{x - 2} \right)}^{\text{3}}}$

We will simplify the given equation.

$\Rightarrow {{\text{x}}^{\text{3}}}\text{ - 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = }{{\text{x}}^{\text{3}}}\text{ - 6}{{\text{x}}^{\text{2}}}\text{ + 12x - 8}$

$\Rightarrow \text{- 4}{{\text{x}}^{\text{2}}}\text{ - x + 1 = - 6}{{\text{x}}^{\text{2}}}\text{ + 12x - 8}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ - 13x + 9 =  0}$

This equation obtained is of the form $\text{a}{{\text{x}}^{\text{2}}}\text{ + bx + c = 0}$.

Therefore, the given equation is a quadratic equation.

2. Represent the following situations in the form of quadratic equations.

(i) The area of a rectangular plot is $\text{528 }{{\text{m}}^{\text{2}}}$. The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.

Ans: It is given that the area of rectangular plot is $\text{528 }{{\text{m}}^{\text{2}}}$.

Let us assume the breadth of the plot be $\text{b = x m}$.

It is given that the length is one more than twice the breadth.

So, the length will be $\text{l = }\left( \text{2x + 1} \right)\text{ m}$.

The area of the rectangle will be  $\text{A = l  }\!\!\times\!\!\text{  b}$

$\therefore \text{l  }\!\!\times\!\!\text{  b = 528}$

$\Rightarrow \text{x}\left( \text{2x + 1} \right)\text{ = 528}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{ + x - 528 = 0}$

Therefore, the required quadratic equation is $\text{2}{{\text{x}}^{\text{2}}}\text{ + x - 528 = 0}$.

(ii) The product of two consecutive positive integers is $\text{306}$. We need to find the integers.

Ans: It is given that the product of two consecutive positive integers is $\text{306}$.

If we assume the first integer to be $\text{x}$, then the next consecutive integer will be $\left( \text{x+1} \right)$.

So, we can write $\text{x}\left( \text{x + 1} \right)\text{ = 306}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + x - 306 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ + x - 306 = 0}$.

(iii) Rohan’s mother is $\text{26}$ years older than him. The product of their ages (in years) $\text{3}$ years from now will be $\text{360}$. We would like to find Rohan’s present age.

Ans: Let us assume Rohan’s age be $\text{x}$.

His mother is $\text{26}$ years older than him.

So, his mother’s age will be $\left( x+26 \right)$

After $\text{3}$ years, Rohan’s age $\text{= x + 3}$

His mother’s age $\text{= x + 26 + 3 = x + 29}$

It is given that after $\text{3}$ years, the product of their ages is $\text{360}$.

$\therefore \left( \text{x + 3} \right)\left( \text{x + 29} \right)\text{ = 360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 32x + 87 = 360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ + 32x - 273 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ + 32x - 273 = 0}$.

(iv) A train travels a distance of $\text{480}$ $\text{km}$ at a uniform speed. If the speed had been $\text{8}$ $\text{km/h}$ less, then it would have taken $\text{3}$ hours more to cover the same distance. We need to find the speed of the train.

Ans: Let us assume the speed of the train be $\text{x km/h}$.

The train travels a distance of $\text{480 km}$ at a uniform speed.

So, the time taken to travel the given distance $\text{= }\frac{\text{480}}{\text{x}}\text{ hrs}$.

It is given that the train would take $\text{3}$ hours more if the

speed had been $\text{8 km/h}$ less.

Now, the new speed of the train $\text{= }\left( \text{x - 8} \right)\text{ km/h}$.

The new time taken to travel the same distance $\text{= }\left( \frac{\text{480}}{\text{x}}\text{ + 3} \right)\text{ hr}$.

We know that $\text{speed  }\!\!\times\!\!\text{  time = distance}$

$\therefore \left( \text{x - 8} \right)\left( \frac{\text{480}}{\text{x}}\text{ + 3} \right)\text{ = 480}$

$\Rightarrow \text{480 + 3x - }\frac{\text{3840}}{\text{x}}\text{ - 24 = 480}$

$\Rightarrow \text{3x - }\frac{\text{3840}}{\text{x}}\text{ = 24}$

$\Rightarrow \text{3}{{\text{x}}^{\text{2}}}\text{ - 24x - 3840 = 0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{ - 8x - 1280 = 0}$

Therefore, the required quadratic equation is ${{\text{x}}^{\text{2}}}\text{ - 8x - 1280 = 0}$.

Conclusion

In Class 10 maths Exercise 4.1, students delve into the basics of quadratic equations. This exercise helps you understand how to identify quadratic equations and distinguish them from other polynomial equations. The key takeaway is recognizing the standard form of a quadratic equation, 𝑎𝑥2+𝑏𝑥+𝑐=0, and understanding the significance of each coefficient. Focus on identifying Quadratic Equations and ensure you can correctly identify equations of the form 𝑎𝑥2+𝑏𝑥+𝑐=0. Mastering these fundamentals is crucial for solving more complex quadratic equations in class 10 Maths ch 4 ex 4.1. Regular practice with Class 10th Exercise 4.1 Solutions will build a strong foundation for further mathematical concepts.

Class 10 Maths Chapter 4: Exercises Breakdown

CBSE Class 10 Maths Chapter 4 Other Study Materials

Chapter-Specific NCERT Solutions for Class 10 Maths

Given below are the chapter-wise NCERT Solutions for Class 10 Maths. Go through these class 10th maths chapter 4 exercise 4.1 solutions to be thoroughly familiar with the concepts.