# NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations Exercise 4.1

## QUADRATIC EQUATIONS EXERCISE 4.1

Q.1: Check whether the following are quadratic equations :

$(\text { i) } \quad(x+1)^{2}=2(x-3) \quad \text { (ii) } x^{2}-2 x=(-2)(3-x)$
$(\text { iii) } \quad(x-2)(x+1)=(x-1)(x+3) \quad \text { (iv) }(x-3)(2 x+1)=x(x+5)$
${ (v) }(2 x-1)(x-3)=(x+5)(x-1) \quad \text { (vi) } x^{2}+3 x+1=(x-2)^{2}$
${ (vii) }(x+2)^{3}=2 x\left(x^{2}-1\right) \quad \text { (viii) } x^{3}-4 x^{2}-x+1=(x-2)^{3}$

## Answers:QUADRATIC EQUATIONS EXERCISE 4.1

 $\begin{array}{l}{\text { (i) } \quad(x+1)^{2}=2(x-3) \Rightarrow x^{2}+2 x+1=2 x-6 \Rightarrow x^{2}+7=0} \\ {\text { It is of the form } a x^{2}+b x+c=0 \text { . }} \\ {\text { Hence, the given equation is a quadratic equation. }}\end{array}$ $\begin{array}{l}{\text { (ii) } \quad x^{2}-2 x=(-2)(3-x) \Rightarrow x^{2}-2 x=-6+2 x \Rightarrow x^{2}-4 x+6=0} \\ {\text { It is of the form } a x^{2}+b x+c=0 \text { . }} \\ {\text { Hence, the given equation is a quadratic equation. }}\end{array}$ $\begin{array}{l}{\text { (iii) }(x-2)(x+1)=(x-1)(x+3) \Rightarrow x^{2}-x-2=x^{2}+2 x-3 \Rightarrow 3 x-1=0} \\ {\text { It is not of the form } a x^{2}+b x+c=0} \\ {\text { Hence, the given equation is not a quadratic equation. }}\end{array}$ $\begin{array}{l}{\text { (iv) } \quad(x-3)(2 x+1)=x(x+5) \Rightarrow 2 x^{2}-5 x-3=x^{2}+5 x \Rightarrow x^{2}-10 x-3=0} \\ {\text { It is of the form } a x^{2}+b x+c=0 \text { . }} \\ {\text { Hence, the given equation is a quadratic equation. }}\end{array}$ $\begin{array}{l}{\text { (v) } \quad(2 x-1)(x-3)=(x+5)(x-1) \Rightarrow 2 x^{2}-7 x+3=x^{2}+4 x-5 \Rightarrow x^{2}-11 x+8=0} \\ {\text { of the form } a x^{2}+b x+c=0 \text { . }} \\ {\text { Hence, the given equation is a quadratic equation. }}\end{array}$ $\begin{array}{l}{\text { (vi) } \quad x^{2}+3 x+1=(x-2)^{2} \Rightarrow x^{2}+3 x+1=x^{2}+4-4 x \Rightarrow 7 x-3=0} \\ {\text { It is not of the form } a x^{2}+b x+c=0} \\ {\text { Hence, the given equation is not a quadratic equation. }} \\ {\text { (vii) } \quad(x+2)^{3}=2 x\left(x^{2}-1\right) \Rightarrow x^{3}+8+6 x^{2}+12 x=2 x^{3}-2 x \Rightarrow x^{3}-14 x-6 x^{2}-8=0} \\ {\text { not of the form } a x^{2}+b x+c=0 \text { . }} \\ {\text { Hence, the given equation is not a quadratic equation. }}\end{array}$ $\begin{array}{l}{\text { (viii) } x^{3}-4 x^{3}-x+1=(x-2)^{3} \Rightarrow x^{3}-4 x^{2}-x+1=x^{3}-8-6 x^{2}+12 x \Rightarrow 2 x^{2}-13 x+9=0} \\ {\text { It is of the form } a x^{2}+b x+c=0} \\ {\text { Hence, the given equation is a quadratic equation. }}\end{array}$
 Q.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.

Answers: Quadratic Equations Exercise 4.1

 $\begin{array}{l}{\text { (i) Let the breadth of the plot be } x \mathrm{m} \text { . }} \\ {\text { Hence, the length of the plot is }(2 x+1) \mathrm{m} \text { . }} \\ {\text { Area of a rectangle = Length } \times \text { Breadth }} \\ {\therefore 528=x(2 x+1)} \\ {\Rightarrow 2 x^{2}+x-528=0}\end{array}$ $\begin{array}{l}{\text { (ii) Let the consecutive integers be } x \text { and } x+1} \\ {\text { It is given that their product is } 306 \text { . }} \\ {\therefore x(x+1)=306 \Rightarrow x^{2}+x-306=0}\end{array}$ $\begin{array}{l}{\text { (iii) Let Rohan’s age be } x \text { . }} \\ {\text { Hence, his mother’s age }=x+26} \\ {3 \text { years hence, }} \\ {\text { Rohan’s age }=x+3} \\ {\text { Mother’s age }=x+26+3=x+29} \\ {\text { It is given that the product of their ages after } 3 \text { years is } 360 \text { . }} \\ {\therefore(x+3)(x+29)=360} \\ {\Rightarrow x^{2}+32 x-273=0}\end{array}$ $\begin{array}{l}{\text { (iv) Let the speed of train be } x \mathrm{km} / \mathrm{h} \text { . }} \\ {\text { Time taken to travel } 480 \mathrm{km}=\frac{480}{x} \text { hrs }} \\ {\text { In second condition, let the speed of train }=(x-8) \mathrm{km} / \mathrm{h}} \\ {\text { It is also given that the train will take } 3 \text { hours to cover the same }} \\ {\text { distance. }} \\ {\text { Therefore, time taken to travel } 480 \mathrm{km}=\left(\frac{480}{x}+3\right)_{\text { hrs }}}\end{array}$ $\begin{array}{l}{\text { Speed } \times \text { Time }=\text { Distance }} \\ {(x-8)\left(\frac{480}{x}+3\right)=480} \\ {\Rightarrow 480+3 x-\frac{3840}{x}-24=480} \\ {\Rightarrow 3 x-\frac{3840}{x}=24} \\ {\Rightarrow 3 x^{2}-24 x+3840=0} \\ {\Rightarrow x^{2}-8 x+1280=0}\end{array}$