Quadratic Equations ⇒⇒ Exercise 4.1

Question 1 (i)

Check whether the following are quadratic equations:
(i) $ (x + 1)^2 = 2(x – 3) $

Solution :

The given equation is $ (x + 1)^2 = 2(x – 3) $

$$ ⇒ { x^2 + 2 x + 1^2 } $$

$$ { = 2x - 6 } $$

( By using the formula for $ (a+b)^2 =a^2 + 2ab+b^2 $ )

$$ ⇒{ x^2 + 2 x + 1 } $$

$$ { - (2x - 6) = 0} $$

$$ ⇒{ x^2 + 2 x + 1 } $$

$$ {- 2x + 6 = 0} $$

$$ ⇒{ x^2 + 7 = 0} $$

Since this equation can be written in the form $ ax^2 + bx + c = 0$, with a = 1, b = 0, and c = 7, and the highest power of x is 2.

Therefore, the given above equation is a quadratic equation.

Question 1 (ii)

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

Solution :

The given equation is $ x ^2 – 2x = (- 2) (3 – x) $

$$ ⇒{ x^2 - 2 x } $$

$$ { = - 6 + 2x } $$

$$ ⇒{ x^2 - 2 x } $$

$$ { - 2x + 6 = 0} $$

$$ { x^2 - 4 x + 6 = 0} $$

Since this equation can be written in the form $ ax^2 + bx + c = 0$, with a = 1, b = -4, and c = 6, and the highest power of x is 2.

Therefore, the given above equation is a quadratic equation.

Question 1 (iii)

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

Solution :

The given equation is $ (x – 2)(x + 1) = (x – 1)(x + 3) $

$$ ⇒{ x^2 + x - 2 x - 2 }$$

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

$$ ⇒{ x^2 - x - 2} $$

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

$$ ⇒{ x^2 - x - 2 } $$

$$ {-(x^2 +2x -3) = 0 } $$

$$ ⇒{ x^2 - x - 2 } $$

$$ { - x^2 - 2x +3 = 0 } $$

$$ ⇒{ - 3x +1 = 0 } $$

$$ ⇒{ 3x - 1 = 0 } $$

Since, The above equation is not in the form of $ ax^2 + bx + c = 0$

Therefore, the given equation is not a Quadratic equation.

Question 1 (iv)

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

Solution :

The given equation is $ (x – 3)(2x + 1) = x(x + 5) $

$$ ⇒{ 2x^2 + x - 6 x - 3 } $$

$$ { = x^2 + 5x } $$

$$ ⇒{ 2x^2 + x - 6 x - 3 } $$

$$ { - (x^2 + 5x) = 0 } $$

$$ ⇒{ 2x^2 - 5x - 3} $$

$$ { - x^2 - 5x = 0 } $$

$$ ⇒{ x^2 - 10x - 3 = 0 } $$

Since this equation can be written in the form $ ax^2 + bx + c = 0$, with a = 1, b = -10, and c = 3, and the highest power of x is 2.

Therefore, the given above equation is a quadratic equation.

Question 1 (v)

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

Solution :

The given equation is $ (2x – 1 )(x – 3) = (x + 5)(x – 1) $

$$ ⇒{ 2x^2 - 6 x - x + 3 } $$

$$ { = x^2 - x + 5x - 5 } $$

$$ ⇒{ 2x^2 - 7 x + 3 } $$

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

$$ ⇒{ 2x^2 - 7 x + 3 } $$

$$ { - (x^2 + 4x - 5 ) = 0 } $$

$$ ⇒{ 2x^2 - 7 x + 3 } $$

$$ { - x^2 - 4x + 5 = 0 } $$

$$ { x^2 - 11x + 8 = 0 } $$

Since this equation can be written in the form $ ax^2 + bx + c = 0$, with a = 1, b = -11, and c = 8, and the highest power of x is 2.

Therefore, the given above equation is a quadratic equation.

Question 1 (vi)

Check whether the following are quadratic equations:
(vi) $ x^2 + 3x + 1 = (x – 2)^2 $

Solution :

The given equation is $ x^2 + 3x + 1 = (x – 2)^2 $

$$ ⇒{ x^2 +3x + 1 } $$

$$ { = x^2 - 4x + 2^2 } $$

( By using the formula for $ (a - b)^2 = a^2 - 2ab + b^2 $ )

$$ ⇒{ x^2 +3x + 1 } $$

$$ {- ( x^2 - 4x + 4 ) = 0 } $$

$$ ⇒ { x^2 +3x + 1 } $$

$$ { - x^2 + 4x - 4 = 0 } $$

$$ ⇒ { 7x - 3 = 0 } $$

Since, The above equation is not in the form of $ ax^2 + bx + c = 0$

Therefore, the given equation is not a Quadratic equation.

Question 1 (vii)

Check whether the following are quadratic equations:
(vii) $(x + 2)^3 = 2x (x^2– 1) $

Solution :

The given equation is $(x + 2)^3 = 2x (x^2– 1) $

$$ { x^3 + (3 × x^2 ×2) } $$

$$ { +(3 × x × 2^2) + 2^3 = 2x^3 - 2x } $$

( By using the formula for $ (a+b)^3 = a^3 + 3a^2b+ 3ab^2 + b^3 $ )

$$ ⇒ { x^3 + 6x^2 + 12x + 8 } $$

$$ { = 2x^3 - 2x } $$

$$ ⇒ {0 = 2x^3 - 2x } $$

$$ {- ( x^3 + 6x^2 + 12x + 8 ) } $$

$$ ⇒ {2x^3 - 2x - x^3 } $$

$$ { - 6x^2 - 12x - 8 = 0 } $$

$$ ⇒ { x^3 - 6x^2 } $$

$$ { - 14x - 8 = 0 } $$

Since, The above equation is not in the form of $ ax^2 + bx + c = 0$

Therefore, the given equation is not a Quadratic equation.

Question 1 (viii)

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

Solution :

The given equation is : $ x^3 – 4x^2 – x + 1 = (x – 2)^3 $

$ ⇒{ x^3 -4x^2 – x + 1 = x^3 } $

$ {- (3 × x^2 ×2) + (3 × x × 2^2) - 2^3 } $

( By using the formula for $ (a - b)^3 = a^3 - 3a^2b+ 3ab^2 - b^3 $ )

$$ ⇒ { x^3 -4x^2 – x + 1 } $$

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

$$ ⇒ { 0 = x^3 - 6 x^2 + 12x - 8 } $$

$$ { - (x^3 -4x^2 – x + 1) } $$

$$ ⇒ { 0 = x^3 - 6 x^2 + 12x - 8 } $$

$$ { - x^3 +4x^2 + x - 1 } $$

$$ ⇒ { 0 = - 2 x^2 } $$

$$ { + 13x - 9 } $$

$$ ⇒ { 2 x^2 - 13x + 9 = 0 } $$

Since this equation can be written in the form $ ax^2 + bx + c = 0$, with a = 2, b = -13, and c = 9, and the highest power of x is 2.

Therefore, the given above equation is a quadratic equation.

Question 2 (i)

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

Solution :

Let the breadth of the plot be x meters.

According to question,

Hence, the length of the plot is = (2x + 1) meters

As we know,

Area of Rectangular Plot = Length × Breadth = $ 528 m^2$

Substituting the expressions for length and breadth into the area formula, we get:,

$$ 528 = (2x + 1)× x $$

$$⇒ 528 = 2 x^2 + x $$

$$ ⇒ 2 x^2 + x - 528 = 0$$

Since this equation can be written in the form $ ax^2 + bx + c = 0$

Therefore, required quadratic equation is $ 2 x^2 + x - 528 = 0$

Question 2 (ii)

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

Solution :

Let's assume the first positive integer is denoted by the variable x

Since the integers are consecutive, the next integer will be x + 1

According to question, given, product of two consecutive integer

$x × (x+1) = 306$

Therefore,

$$ x × (x + 1) =306 $$

$$⇒ x^2 + x = 306 $$

$$ ⇒ x^2 + x - 306 = 0$$

Since this equation can be written in the form $ ax^2 + bx + c = 0$

Therefore, required quadratic equation is $ x^2 + x - 306 = 0$

Question 2 (iii)

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

Solution :

Let, present age of Rohan = x years

then, present age of Rohan’s mother be (x + 26) years

3 years from now,

Rohan's age = (x + 3)years

Age of Rohan’s mother will be = (x + 26 +3) = (x + 29) years

The product of their ages 3 years from now will be 360 so that :

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

$$⇒ x^2 + 29x + 3x + 87 = 360 $$

$$ ⇒ x^2 + 32x + 87 – 360 = 0$$

$$ ⇒ x^2 + 32x – 273 = 0$$

Since this equation can be written in the form $ ax^2 + bx + c = 0$

Therefore, required quadratic equation is $ x^2 + 32x – 273 = 0$

Question 2 (iv)

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

Solution :

Let us consider, The speed of the train = x km/h

As we know, Time = Distance / Speed

Time taken by the train to cover 480 km = $ 480 \over x $ km/h

In second condition, let the speed of train = (x – 8) km/h

Time taken by the train to cover 480 km = $ {480 \over (x – 8) } $ km/h

We know that, difference in time = 3 hrs

$$ {480 \over (x – 8) } - {480 \over x} = 3 $$

$$⇒ {{(480x - 480x + 3840 )} \over (x^2 – 8x) } = 3 $$

$$⇒ {3840 = 3 × (x^2 – 8x) }$$

$$⇒ {3840 = 3 x^2 – 24x }$$

$$⇒ { 3 x^2 – 24x - 3840 = 0 }$$

dividing by 3

$$⇒ { x^2 – 8x - 1280 = 0 }$$

Since this equation can be written in the form $ ax^2 + bx + c = 0$

Therefore, required quadratic equation is $ { x^2 – 8x - 1280 = 0 }$