NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

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

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

Ans : \(\begin{aligned}(i) & x^{2}-3 x-10 \\ &=x^{2}-5 x+2 x-10 \\ &=x(x-5)+2(x-5) \\ &=(x-5)(x+2) \end{aligned}\)
\( \begin{array}{l}{\text { Roots of this equation are the values for which }(x-5)(x+2)=0} \\ {\therefore x-5=0 \text { or } x+2=0} \\ {\text { i.e., } x=5 \text { or } x=-2}\end{array}\)


\(\begin{aligned}(\text { ii) }& 2 x^{2}+x-6 \\ &=2 x^{2}+4 x-3 x-6 \\ &=2 x(x+2)-3(x+2) \\ &=(x+2)(2 x-3) \end{aligned}\)
\(\begin{array}{l}{\text { Roots of this equation are the values for which }(x+2)(2 x-3)=0} \\ {\therefore x+2=0 \text { or } 2 x-3=0} \\ {\text { i.e., } x=-2 \text { or } x=\frac{3}{2}}\end{array}\)


\(\begin{aligned} \text { (iii) } & \sqrt{2} x^{2}+7 x+5 \sqrt{2} \\ &=\sqrt{2} x^{2}+5 x+5 x+5 \sqrt{2} \\ &=x(\sqrt{2} x+5)+\sqrt{2}(\sqrt{2} x+5) \\ &=(\sqrt{2} x+5)(x+\sqrt{2}) \end{aligned}\)
\( \begin{array}{l}{\text { Roots of this equation are the values for which }(\sqrt{2} x+5)(x+\sqrt{2})=0} \\ {\therefore \sqrt{2} x+5=0 \text { or } x+\sqrt{2}=0} \\ {\text { i.e., } x=\frac{-5}{\sqrt{2}} \text { or } x=-\sqrt{2}}\end{array}\)


(iv) 
\(\begin{aligned} & 2 x^{2}-x+\frac{1}{8} \\ &=\frac{1}{8}\left(16 x^{2}-8 x+1\right) \\ &=\frac{1}{8}\left(16 x^{2}-4 x-4 x+1\right) \\ &=\frac{1}{8}(4 x(4 x-1)-1(4 x-1)) \\ &=\frac{1}{8}(4 x-1)^{2} \end{aligned}\)
\(\begin{array}{l}{\text { Roots of this equation are the values for which }(4 x-1)^{2}=0} \\ {\text { Therefore, }(4 x-1)=0 \text { or }(4 x-1)=0} \\ {\text { i.e., } x=\frac{1}{4} \text { or } x=\frac{1}{4}}\end{array}\)


\(\begin{aligned}(v) & 100 x^{2}-20 x+1 \\ &=100 x^{2}-10 x-10 x+1 \\ &=10 x(10 x-1)-1(10 x-1) \\ &=(10 x-1)^{2} \end{aligned}\)
\(\begin{array}{l}{\text { Roots of this equation are the values for which }(10 x-1)^{2}=0} \\ {\text { Therefore, }(10 x-1)=0 \text { or }(10 x-1)=0} \\ {\text { i.e., } x=\frac{1}{10} \text { or } x=\frac{1}{10}}\end{array}\) 

Ans : \(\begin{array}{l}{\text { (i) Let the number of John's marbles be } x \text { . }} \\ {\text { The refore, number of Jivanti's marble }=45-x} \\ {\text { After losing } 5 \text { marbles, }} \\ {\text { Number of John's marbles = } x-5} \\ {\text { Number of Jivanti's marbles = } 45-x-5=40-x} \\ {\text { It is given that the product of their marbles is } 124 .}\end{array}\)
\(\begin{array}{l}{\therefore(x-5)(40-x)=124} \\ {\Rightarrow x^{2}-45 x+324=0} \\ {\Rightarrow x^{2}-36 x-9 x+324=0} \\ {\Rightarrow x(x-36)-9(x-36)=0} \\ {\Rightarrow(x-36)(x-9)=0}\end{array}\)
\(\begin{array}{l}{\text { Either } x-36=0 \text { or } x-9=0} \\ {\text { i.e., } x=36 \text { or } x=9} \\ {\text { If the number of John's marbles }=36 \text { , }} \\ {\text { Then, number of Jivanti's marbles }=45-36=9} \\ {\text { If number of John's marbles }=9,} \\ {\text { Then, number of Jivanti's marbles }=45-9=36}\end{array}\)


\(\begin{array}{l}{\text { (ii) Let the number of toys produced be } x \text { . }} \\ {\therefore \text { cost of production of each toy }=R s(55-x)} \\ {\text { It is given that, total production of the toys }=Rs 750}\end{array}\)
\(\begin{array}{l}{\therefore x(55-x)=750} \\ {\Rightarrow x^{2}-55 x+750=0} \\ {\Rightarrow x^{2}-25 x-30 x+750=0} \\ {\Rightarrow x(x-25)-30(x-25)=0} \\ {\Rightarrow(x-25)(x-30)=0}\end{array}\)
\( \begin{array}{l}{\text { Either } x-25=0 \text { or } x-30=0} \\ {\text { i.e., } x=25 \text { or } x=30} \\ {\text { Hence, the number of toys will be either } 25 \text { or } 30 \text { . }}\end{array}\) 

Ans : \(\begin{array}{l}{\text { Let the first number be } x \text { and the second number is } 27-x \text { . }} \\ {\text { Therefore, their product }=x(27-x)} \\ {\text { It is given that the product of these numbers is } 182 \text { . }} \\ {\text { Therefore, } x(27-x)=182}\end{array}\)
\(\begin{array}{l}{\Rightarrow x^{2}-27 x+182=0} \\ {\Rightarrow x^{2}-13 x-14 x+182=0} \\ {\Rightarrow x(x-13)-14(x-13)=0} \\ {\Rightarrow(x-13)(x-14)=0}\end{array}\)
\(\begin{array}{l}{\text { Either } x-13=0 \text { or } x-14=0} \\ {\text { l.e., } x=13 \text { or } x=14} \\ {\text { If first number }=13 \text { , then }} \\ {\text { other number }=27-13=14} \\ {\text { If first number }=14, \text { then }} \\ {\text { If first number }=27-14=13} \\ {\text { Therefore, the numbers are } 13 \text { and } 14 .}\end{array}\)