NCERT Solutions for Class 10 Maths Chapter 4 Exercise 4.1 - 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: $ax^2+bx+c = 0$.

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: $ax^2+bx+c = 0,a\neq0$.

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: $ax^2+bx+c = 0,a\neq0$.

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: $ax^2+bx+c = 0, a \neq 0$.

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: $ax^2+bx+c = 0, a\neq0$.

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: $ax^2+bx+c = 0$.

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: $ax^2+bx+c = 0$.

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 $'b'$.

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 $'x'$ and $'x+1'$.

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 $'x'$ years.

Then his mother's age will be: $'x+26'$ years.

After three years,

Rohan's age will be $'x+3'$ years, and his mother's age will be $'x+29'$ 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 $'s'$ km/h.

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

$\therefore$ The time taken will be

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

If the speed had been $8\ km/h$ 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 $x^2 -8x-1280 = 0$