Exercise 4.1 (Revised) - Chapter 4 - Quadratic Equations - Ncert Solutions class 10 - Maths

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NCERT Solutions for Class 10 Maths Chapter 4: Quadratic Equations

Ex 4.1 Question 1.

Check whether the following are Quadratic Equations.
(i) $(x+1)^2=2(x-3)$
(ii) $x^2-2 x=(-2)(3-x)$
(iii) $(x-2)(x+1)=(x-1)(x+3)$
(iv) $(x-3)(2 x+1)=x(x+5)$
(v) $(2 x-1)(x-3)=(x+5)(x-1)$
(vi) $x^2+3 x+1=(x-2)^2$
(vii) $(x+2)^3=2 x\left(x^2-1\right)$
(viii) $x^3-4 x^2-x+1=(x-2)^3$

Answer.

(i) $(x+1)^2=2(x-3)$
$
\begin{aligned}
& \left\{(a+b)^2=a^2+2 a b+b^2\right\} \\
& \Rightarrow x^2+1+2 x=2 x-6 \\
& \Rightarrow x^2+7=0
\end{aligned}
$

Here, degree of equation is 2 .

Therefore, it is a Quadratic Equation.
(ii) $x^2-2 x=(-2)(3-x)$

$
\begin{aligned}
& \Rightarrow x^2-2 x=-6+2 x \\
& \Rightarrow x^2-2 x^{-2 x+6}=0 \\
& \Rightarrow x^2-4 x+6=0
\end{aligned}
$

Here, degree of equation is 2 .
Therefore, it is a Quadratic Equation.
$
\begin{aligned}
& \text { (iii) }(x-2)(x+1)=(x-1)(x+3) \\
& \Rightarrow x^2+x-2 x-2=x^2+3 x-x-3=0 \\
& \Rightarrow x^2+x-2 x-2-x^2-3 x+x+3=0 \\
& \Rightarrow x-2 x-2-3 x+x+3=0 \\
& \Rightarrow-3 x+1=0
\end{aligned}
$

Here, degree of equation is 1 .
Therefore, it is not a Quadratic Equation.
$
\begin{aligned}
& \text { (iv) }(x-3)(2 x+1)=x(x+5) \\
& \Rightarrow 2 x^2+x-6 x-3=x^2+5 x \\
& \Rightarrow 2 x^2+x-6 x-3-x^2-5 x=0 \\
& \Rightarrow x^2-10 x-3=0
\end{aligned}
$

Here, degree of equation is 2 .
Therefore, it is a quadratic equation.
(v) $(2 x-1)(x-3)=(x+5)(x-1)$

$\begin{aligned}
& \Rightarrow 2 x^2-6 x-x+3=x^2-x+5 x-5 \\
& \Rightarrow 2 x^2-7 x+3-x^2+x-5 x+5=0
\end{aligned}$

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

Here, degree of Equation is 2.
Therefore, it is a Quadratic Equation.
$
\begin{aligned}
& \text { (vi) } x^2+3 x+1=(x-2)^2 \\
& \left\{(a-b)^2=a^2-2 a b+b^2\right\} \\
& \Rightarrow x^2+3 x+1=x^2+4-4 x \\
& \Rightarrow x^2+3 x+1-x^2+4 x-4=0 \\
& \Rightarrow 7 x-3=0
\end{aligned}
$

Here, degree of equation is 1.
Therefore, it is not a Quadratic Equation.
$
\begin{aligned}
& \text { (vii) }(x+2)^3=2 x\left(x^2-1\right) \\
& \left\{(a+b)^3=a^3+b^3+3 a b(a+b)\right\} \\
& \Rightarrow x^3+2^3+3(x)(2)(x+2)=2 x\left(x^2-1\right) \\
& \Rightarrow x^3+8+6 x(x+2)=2 x^3-2 x \\
& \Rightarrow 2 x^3-2 x-x^3-8-6 x^2-12 x=0 \\
& \Rightarrow x^3-6 x^2-14 x-8=0
\end{aligned}
$

Here, degree of Equation is 3.

Therefore, it is not a quadratic Equation.
$
\text { (viii) } x^3-4 x^2-x+1=(x-2)^3
$

$
\begin{aligned}
& \left\{(a-b)^3=a^3-b^3-3 a b(a-b)\right\} \\
& \Rightarrow x^3-4 x^2-x+1=x^3-2^3-3(x)(2)(x-2) \\
& \Rightarrow-4 x^2-x+1=-8-6 x^2+12 x \\
& \Rightarrow 2 x^2-13 x+9=0
\end{aligned}
$

Here, degree of Equation is 2.
Therefore, it is a Quadratic Equation.

Ex 4.1 Question 2.

Represent the following situations in the form of Quadratic Equations:
(i) The area of rectangular plot is $528 \mathrm{~m}^2$. 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 numbers 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) after 3 years will be 360 . We would like to find Rohan's present age.
(iv) A train travels a distance of $480 \mathrm{~km}$ at 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 speed of the train.

Answer.

(i) We are given that area of a rectangular plot is $528 \mathrm{~m}^2$.
Let breadth of rectangular plot be $\mathrm{x}$ metres
Length is one more than twice its breadth.
Therefore, length of rectangular plot is $(2 x+1)$ metres
Area of rectangle $=$ length $\times$ breadth
$
\begin{aligned}
& \Rightarrow 528=x(2 x+1) \\
& \Rightarrow 528=2 x^2+x
\end{aligned}
$

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

This is required Quadratic Equation.
(ii) Let two consecutive numbers be $\mathrm{x}$ and $(\mathrm{x}+1)$.

It is given that $x(x+1)=306$
$
\begin{aligned}
& \Rightarrow x^2+x=306 \\
& \Rightarrow x^2+x^{-306}=0
\end{aligned}
$

This is the required Quadratic Equation.
(iii) Let present age of Rohan $=\mathrm{x}$ years

Let present age of Rohan's mother $=(x+26)$ years
Age of Rohan after 3 years $=(x+3)$ years
Age of Rohan's mother after 3 years $=x+26+3=(x+29)$ years
According to given condition:
$
\begin{aligned}
& (x+3)(x+29)=360 \\
& \Rightarrow x^2+29 x+3 x+87=360 \\
& \Rightarrow x^2+32 x-273=0
\end{aligned}
$

This is the required Quadratic Equation.
(iv) Let speed of train be $x \mathrm{~km} / \mathrm{h}$

Time taken by train to cover $480 \mathrm{~km}=\frac{480}{x}$ hours
If, speed had been $8 \mathrm{~km} / \mathrm{h}$ less then time taken would be $\frac{480}{x-8}$ hours

According to given condition, if speed had been $8 \mathrm{~km} / \mathrm{h}$ less then time taken is 3 hours less.
Therefore, $\frac{480}{x-8}-\frac{480}{x}=3$

$
\begin{aligned}
& \Rightarrow 480(x-x+8)=3 x(x-8) \\
& \Rightarrow 3840=3 x^2-24 x \\
& \Rightarrow 3 x^2-24 x-3840=0
\end{aligned}
$

Dividing equation by 3 , we get
$
\Rightarrow x^2-8 x-1280=0
$

This is the required Quadratic Equation.