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

Q1 (ii) Check whether the following are quadratic equations: $x^2-2x=(-2)(3-x)$

Answer:

Given equation $x^2-2x=(-2)(3-x)$ can be written as:

$\Rightarrow x^2-2x=-6+2x$

i.e., $x^2-4x+6=0$

This equation is of the type: $a{x}^2+bx+c=0$.

Hence, the given equation is a quadratic equation.

Q1 (iii) Check whether the following are quadratic equations: $(x-2)(x+1)=(x-1)(x+3)$

Answer:

L.H.S. $(x-2)(x+1)$ can be written as:

$=x^2+x-2x-2=x^2-x-2$

and R.H.S $(x-1)(x+3)$ can be written as:

$=x^2+3x-x-3=x^2+2x-3$

$\Rightarrow x^2-x-2=x^2+2x-3$

i.e., $3x-1=0$

The equation is of the type: $a{x}^2+bx+c=0,a\ne 0$.

Hence, the given equation is not a quadratic equation since a = 0.

Q1 (iv) Check whether the following are quadratic equations: $(x-3)(2x+1)=x(x+5)$

Answer:

L.H.S. $(x-3)(2x+1)$ can be written as:

$=2x^2+x-6x-3=2x^2-5x-3$

and R.H.S $(x)(x+5)$ can be written as:

$=x^2+5x$

$\Rightarrow 2x^2-5x-3=x^2+5x$

i.e., $x^2-10x-3=0$

This equation is of type: $a{x}^2+bx+c=0,a\ne 0$.

Hence, the given equation is a quadratic equation.

Q1 (v) Check whether the following are quadratic equations: $(2x-1)(x-3)=(x+5)(x-1)$

Answer:

L.H.S. $(2x-1)(x-3)$ can be written as:

$=2x^2-6x-x+3=2x^2-7x+3$

and R.H.S $(x+5)(x-1)$ can be written as:

$=x^2-x+5x-5=x^2+4x-5$

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

i.e., $x^2-11x+8=0$

This equation is of type: $a{x}^2+bx+c=0,a\ne 0$.

Hence, the given equation is a quadratic equation.

Q1 (vi) Check whether the following are quadratic equations: $x^2+3x+1=(x-2{)}^2$

Answer:

L.H.S. $x^2+3x+1$

and R.H.S $(x-2{)}^2$ can be written as:

$=x^2-4x+4$

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

i.e., $7x-3=0$

This equation is NOT of type: $a{x}^2+bx+c=0,a\ne 0$.

Here, a = 0, hence, the given equation is not a quadratic equation.

Q1 (vii) Check whether the following are quadratic equations:$(x+2{)}^3=2x(x^2-1)$

Answer:

L.H.S. $(x+2{)}^3$ can be written as:

$=x^3+8+6x(x+2)=x^3+6x^2+12x+8$

and R.H.S $2x(x^2-1)$ can be written as:

$=2x^3-2x$

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

i.e., $x^3-6x^2-14x-8=0$

This equation is NOT of type: $a{x}^2+bx+c=0$.

Hence, the given equation is not a quadratic equation.

Q1 (viii) Check whether the following are quadratic equations: $x^3-4x^2-x+1=(x-2{)}^3$

Answer:

L.H.S. $x^3-4x^2-x+1$ ,

and R.H.S $(x-2{)}^3$ can be written as:

$=x^3-6x^2+12x-8$

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

i.e., $2x^2-13x+9=0$

This equation is of the type: $a{x}^2+bx+c=0$.

Hence, the given equation is a quadratic equation.

Q2 (i) Represent the following situations in the form of quadratic equations: The area of a rectangular plot is $528m^2$. 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.

Answer:

Given that the area of a rectangular plot is $528m^2$.

Let the breadth of the plot be ${}^{\prime }{b}^{\prime }$.

Then, the length of the plot will be: $=2b+1$.

Therefore, the area will be:

$=b(2b+1)m^2$ which is equal to the given plot area $528m^2$.

$\Rightarrow 2b^2+b=528$

$\Rightarrow 2b^2+b-528=0$

Hence, the length and breadth of the plot will satisfy the equation $2b^2+b-528=0$

Q2 (ii) Represent the following situations in the form of quadratic equations:The product of two consecutive positive integers is 306. We need to find the integers.

Answer:

Given that the product of two consecutive integers is $306.$

Let two consecutive integers be ${}^{\prime }{x}^{\prime }$ and ${}^{\prime }x+1^{\prime }$.

Then, their product will be:

$x(x+1)=306$

Or $x^2+x-306=0$ .

Hence, the two consecutive integers will satisfy this quadratic equation $x^2+x-306=0$.

Q2 (iii) Represent the following situations in the form of quadratic equations: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 the age of Rohan be ${}^{\prime }{x}^{\prime }$ years.

Then his mother's age will be: ${}^{\prime }x+{26}^{\prime }$ years.

After three years,

Rohan's age will be ${}^{\prime }x+3^{\prime }$ years, and his mother's age will be ${}^{\prime }x+{29}^{\prime }$ years.

Then, according to the question,

The product of their ages 3 years from now will be:

$\Rightarrow (x+3)(x+29)=360$

$\Rightarrow x^2+3x+29x+87=360$ Or

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

Hence, the age of Rohan satisfies the quadratic equation $x^2+32x-273=0$.

Q2 (iv) Represent the following situations in the form of quadratic equations: 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.

Answer:

Let the speed of the train be ${}^{\prime }{s}^{\prime }$ km/h.

The distance to be covered by the train is $480km$.

$\therefore$ The time taken will be

$=\frac{480}{s}hours$

If the speed had been $8km/h$ less, the time taken would be: $\frac{480}{s-8}hours$ .

Now, according to the question,

$\frac{480}{s-8}-\frac{480}{s}=3$

$\Rightarrow \frac{480x-480(x-8)}{(x-8)x}=3$

$\Rightarrow 480x-480x+3840=3(x-8)x$

$\Rightarrow 3840=3x^2-24x$

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

Dividing by 3 on both sides

$x^2-8x-1280=0$

Hence, the speed of the train satisfies the quadratic equation $x^2-8x-1280=0$