NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 || CBSE Class 10 maths chapter 4 Ex 4.1

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NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 || CBSE Class 10 maths chapter 4 Ex 4.1
Exercise 4.1

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 || CBSE Class 10 maths chapter 4 Ex 4.1

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1. Check whether the following are quadratic equations :

(i) $ (x+1)^{2}=2(x-3) $

Answer

For a equation to be quadratic equation, degree of the equation(highest power of the variable in the equation) should be 2 Thus a quadratic equation is of the form, $ \mathrm{ax}^{2}+b x+c=0 $, where $ a \neq 0 $
(i) $ (\mathrm{x}+1)^{2}=2(\mathrm{x}-3) $
We know $ (a+b)^{2}=a^{2}+b^{2}+2\ a b $
$\Rightarrow \mathrm{x}^{2}+2 \mathrm{x}+1=2 \mathrm{x}-6$
$\Rightarrow \mathrm{x}^{2}+2 \mathrm{x}+1-2 \mathrm{x}+6=0$
$\Rightarrow \mathrm{x}^{2}+7=0$
Degree (highest power) of the equation is 2
Equation is in the form of $ \mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}=0 $, with $ \mathrm{a} \neq 0 $
So, we can say that the given equation is a quadratic equation.

(ii) $ x^{2}-2 x=(-2)(3-x) $

Answer

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 || CBSE Class 10 maths chapter 4 Ex 4.1

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(iii) $ (\mathrm{x}-2)(\mathrm{x}+1)=(\mathrm{x}-1)(\mathrm{x}+3) $

Answer

For a equation to be quadratic equation, degree of the equation(highest power of the variable in the equation) should be 2 Thus a quadratic equation is of the form, $ \mathrm{ax}^{2}+b x+c=0 $, where $ a \neq 0 $
(iii) $(\mathrm{x}-2)(\mathrm{x}+1)=(\mathrm{x}-1) (\mathrm{x}+3)$
$\mathrm{x}(\mathrm{x}+1)-2 (\mathrm{x}+1)=\mathrm{x} (\mathrm{x}+3)-1 (\mathrm{x}+3)$
$\Rightarrow \mathrm{x}^{2}+\mathrm{x}-2 \mathrm{x}-2=\mathrm{x}^{2}+3 \mathrm{x}-\mathrm{x}-3$
$\Rightarrow \mathrm{x}^{2}-\mathrm{x}-2=\mathrm{x}^{2}+2 \mathrm{x}-3$
$\Rightarrow \mathrm{x}^{2}-\mathrm{x}-2-\mathrm{x}^{2}-2 \mathrm{x}+3=0$
$\Rightarrow-3 \mathrm{x}+1=0$
$\Rightarrow 3 \mathrm{x}-1=0$
Degree (highest power) of the equation is 1
Equation is not in the form $ a x^{2}+b x+c=0 $.
Hence, the given equation is not a quadratic equation.

(iv) $ (x-3)(2 x+1)=x(x+5) $

Answer

For a equation to be quadratic equation, degree of the equation(highest power of the variable in the equation) should be 2 Thus a quadratic equation is of the form, $ \mathrm{ax}^{2}+b x+c=0 $, where $ a \neq 0 $
(iv)$ (x-3)(2 x+1)=x(x+5) $ $\Rightarrow x(2 x+1)-3(2 x+1)=x^{2}+5 x$
$\Rightarrow 2 x^{2}+x-6 x-3=x^{2}+5 x$
$\Rightarrow 2 x^{2}-5 x-3=x^{2}+5 x$
$\Rightarrow 2 x^{2}-5 x-3-x^{2}-5 x=0$
$\Rightarrow x^{2}-10 x-3=0$
Degree of the equation is 2
Equation is in the form $ a^{2}+b x+c=0 $, with $ a \neq 0 $
Hence, the given equation is a quadratic equation.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 || CBSE Class 10 maths chapter 4 Ex 4.1

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(v) $ (2 x-1)(x-3)=(x+5)(x-1) $

Answer

For a equation to be quadratic equation, degree of the equation(highest power of the variable in the equation) should be 2 Thus a quadratic equation is of the form, $ \mathrm{ax}^{2}+b x+c=0 $, where $ a \neq 0 $
(v) $(2 \mathrm{x}-1)(\mathrm{x}-3)=(\mathrm{x}+5)(\mathrm{x}-1)$
$\Rightarrow 2 \mathrm{x}(\mathrm{x}-3)-1(\mathrm{x}-3)=\mathrm{x}(\mathrm{x}-1)+5(\mathrm{x}-1)$
$\Rightarrow 2 \mathrm{x}^{2}-6 \mathrm{x}-\mathrm{x}+3=\mathrm{x}^{2}-\mathrm{x}+5 \mathrm{x}-5$
$\Rightarrow 2 \mathrm{x}^{2}-7 \mathrm{x}+3=\mathrm{x}^{2}+4 \mathrm{x}-5$
$\Rightarrow 2 \mathrm{x}^{2}-7 \mathrm{x}+3-\mathrm{x}^{2}-4 \mathrm{x}+5=0$
$\Rightarrow \mathrm{x}^{2}-11 \mathrm{x}+8=0$
Degree of the equation is 2
Equation is in the form $ a x^{2}+b x+c=0 $, with $ a \neq 0 $
Hence, the given equation is a quadratic equation.

(vi) $ x^{2}+3 x+1=(x-2)^{2} $

Answer

For a equation to be quadratic equation, degree of the equation(highest power of the variable in the equation) should be 2 Thus a quadratic equation is of the form, $ \mathrm{ax}^{2}+b x+c=0 $, where $ a \neq 0 $
$\text { (vi) } x^{2}+3 \mathrm{x}+1=(\mathrm{x}-2)^{2}$
$\Rightarrow \mathrm{x}^{2}+3 \mathrm{x}+1=\mathrm{x}^{2}+4-4 \mathrm{x}$
$\Rightarrow \mathrm{x}^{2}+3 \mathrm{x}+1-\mathrm{x}^{2}-4+4 \mathrm{x}=0$
$\Rightarrow 7 \mathrm{x}-3=0$
Degree of the equation is 1
It is not of the form $ a x^{2}+b x+c=0 $.
Hence, the given equation is not a quadratic equation.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 || CBSE Class 10 maths chapter 4 Ex 4.1

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

Answer

For a equation to be quadratic equation, degree of the equation(highest power of the variable in the equation) should be 2 Thus a quadratic equation is of the form, $ \mathrm{ax}^{2}+b x+c=0 $, where $ a \neq 0 $
(vii) $ (\mathrm{x}+2)^{3}=2 \mathrm{x}\left(\mathrm{x}^{2}-1\right) $
Apply the formula $ (a+b)^{3}=a^{3}+b^{3}+3 a b(a+b) $
$ \Rightarrow x^{3}+8+6 x^{2}+12 x=2 x^{3}-2 x $
$ \Rightarrow x^{3}+8+6 x^{2}+12 x-2 x^{3}+2 x=0$
$\Rightarrow-x^{3}+8+6 x^{2}+14 x=0$
$\Rightarrow x^{3}-14 x-6 x^{2}-8=0$
Degree of the equation is 3
It is not of the form $ a x^{2}+b x+c=0 $.
Hence, the given equation is not a quadratic equation.

(viii) $ x^{3}-4 x^{2}-x+1=(x-2)^{3} $

Answer

For a equation to be quadratic equation, degree of the equation(highest power of the variable in the equation) should be 2 Thus a quadratic equation is of the form, $ \mathrm{ax}^{2}+b x+c=0 $, where $ a \neq 0 $
(viii) $ x^{3}-4 x^{2}-x+1=(x-2)^{3}$
$(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}$
$\Rightarrow x^{3}-4 x^{2}-x+1=x^{3}-8-6 x^{2}+12 x$
$\Rightarrow x^{3}-4 x^{2}-x+1-x^{3}+8+6 x^{2}-12 x=0$
$\Rightarrow 2 x^{2}-13 x+9=0$
Degree of the equation is 2 and
It is of the form $ a x^{2}+b x+c=0 $, with $ a \neq 0 $
Hence, the given equation is a quadratic equation.

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 || CBSE Class 10 maths chapter 4 Ex 4.1

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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.

Answer

The area of a rectangular plot is 528 m2 .
Let the breadth of the plot be x m .
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.
Thus, the length of the plot is $ (2 x+1) \mathrm{m} $.
Area of a rectangle $ = $ Length $ \times $ Breadth
$ \therefore 528=\mathrm{x}(2 \mathrm{x}+1) $
$ \Rightarrow 2 \mathrm{x}^{2}+\mathrm{x}-528=0 $ (required quadratic form)

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

Answer

The product of two consecutive positive integers is 306. We need to find the integers.
Let the consecutive integers be x and $ \mathrm{x}+1 $.
It is given that their product is 306.
$ \therefore \mathrm{x}(\mathrm{x}+1)=306 $
$ \Rightarrow \mathrm{x}^{2}+\mathrm{x}-306=0 $ (required quadratic form)

(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.

Answer

Let Rohan's present age be x .
Given, Rohan's Mother is 26 years older than him
Hence, his mother's age $ =x+26 $
3 years hence,
Rohan's age $ =\mathrm{x}+3 $
Mother's age $ =\mathrm{x}+26+3 $
$ =\mathrm{x}+29 $
It is given that the product of their ages after 3 years is 360.
$ \therefore(\mathrm{x}+3)(\mathrm{x}+29)=360 $
$ x^{2}+3 x+29 x+87=360 $
$ \Rightarrow \mathrm{x}^{2}+32 \mathrm{x}-273=0 $ (required quadratic form)

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 || CBSE Class 10 maths chapter 4 Ex 4.1

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

Answer

Let the speed of train be $ x \mathrm{~km} / \mathrm{h} $.
As speed $ = $ distance $ / $ time
$ \Rightarrow $ Time taken for travel $ 480 \mathrm{~km}=\frac{480}{x} h r s $
In the second condition,
speed of train $ =(\mathrm{x}-8) \mathrm{km} / \mathrm{h} $
Given that the train will take 3 hours more to cover the same distance.
Therefore, Time taken for traveling $ 480 \mathrm{~km}=\left(\frac{480}{x}+3\right) \mathrm{hrs} $
Speed $ \times $ Time $ = $ Distance
$ (x-8)\left(\frac{480}{x}+3\right)=480$
$\Rightarrow 480+3 x-\frac{3840}{x}-24=480$
$\Rightarrow 3 x-\frac{3840}{x}=24$
$\Rightarrow 3 \mathrm{x}^{2}-3840=24 \mathrm{x}$
$\Rightarrow 3 \mathrm{x}^{2}-24 \mathrm{x}-3840=0$
$\Rightarrow \mathrm{x}^{2}-8 \mathrm{x}-1280=0 \text { (required quadratic form) }$

NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 || CBSE Class 10 maths chapter 4 Ex 4.1

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NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 || CBSE Class 10 maths chapter 4 Ex 4.1

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