class-10th-Quadratic Equation 4.1

1. Check whether the following are Quadratic Equations.

(i)  (x+1)²= 2 (x − 3)

(ii)  x²-2x= (−2) (3 − x)

(iii) (x − 2) (x + 1) = (x − 1) (x + 3)

(iv) (x − 3) (2x + 1) = x (x + 5)

(v) (2x − 1) (x − 3) = (x + 5) (x − 1)

(vi)x²+3x+1=(x-2)²

(vii)(x+2)³=2x(x²-1)

(viii)x³-4x²-x+1=(x-2)²

 

Ans. (i)(x+1)²=2(x-3) 

{(a+b)²=a²+2ab+b²}

⇒ x²+1+2x=2x-6

⇒ x²+7=0

Here, degree of equation = 2.

Therefore, it is a Quadratic Equation.

(ii) x²-2x=(-2)(3-x)

⇒ x²-2x=(-2)(3-x)

⇒ x²-2x=-6+2x

⇒ x²-2x+6-2x=0

⇒ x²-4x+6=0

Here, degree of equation is=2.

Therefore, it is a Quadratic Equation.

(iii) (x − 2) (x + 1) = (x − 1) (x + 3)

⇒ x²+x-2x-2=x²+3x-x-3=0

⇒ x²+x-2x-2-x²-3x+x+3=0

⇒ x-2x-2-3x+x+3=0

⇒ -3x+1=0

Here, degree of equation = 1.

Therefore, it is not a Quadratic Equation.

(iv) (x − 3) (2x + 1) = x (x + 5)

⇒ 2x²+x-6x-3=x²+5x

⇒ 2x²+x-6x-3=x²+5x

⇒ 2x²+x-6x-3-x²-5x=0

⇒ x²-10x-3=0

Here, degree of equation=2.

Therefore, it is a quadratic equation.

(v) (2x − 1) (x − 3) = (x + 5) (x − 1)

⇒ 2x²-6x-x+3= x²-x+5x-5

⇒ 2x²-6x-x+3-x²+x-5x+5=0

⇒ x²-11x+8=0

⇒ x² − 11x + 8 = 0

Here, degree of Equation=2.

Therefore, it is a Quadratic Equation.

(vi) x²+3x+1=(x-2)²

{(a-b)²=a²-2ab+b²}

⇒ x²+3x+1=x² +4-4x

⇒ x² +3x+1-x² -4+4x=0

⇒ 7x-3=0

Here, degree of equation=1.

Therefore, it is not a Quadratic Equation.

(vii)(x+2)³=2x(x²-1)

{(a+b)³=a³+b³+3ab(a+b)}

⇒ x³+2³+3(x)(2)(x+2)=2x(x²-1)

⇒ x³+8+6x(x+2)=2x(x²-1)

⇒ x³+8+6x²+12x=2x³-2x

⇒ x³+8+6x²+12x-2x³+2x=0

⇒ -x³+6x²+14x+8=0

⇒ x³-6x²-14x-8=0

Here, degree of Equation=3.

Therefore, it is not a quadratic Equation.

(viii) x³-4x²-x-1=(x-2)³

{(a-b)³=a³-b³-3ab(a-b)}

⇒ x³-4x²-x+1=x³-2³-3(x)(2)(x-2)

⇒ x³-4x²-x+1=x³-8-6x²+12x

⇒ x³-4x²-x+1-x³+8+6x²-12x=0

⇒ 2x²-13x+9=0

Here, degree of Equation = 2.

Therefore, it is a Quadratic Equation.

2. Represent the following situations in the form of Quadratic Equations:

(i) The area of rectangular plot is 528m². 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 positive integers 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 from now will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find speed of the train.

Ans. (i) We are given that area of a rectangular plot is 528m².

Let breadth of rectangular plot be x metres

Length is one more than twice its breadth.

Therefore, length of rectangular plot is (2x + 1) metres

Area of rectangle = length × breadth

⇒ 528 = x (2x + 1)

⇒ 528 = 2x²+x 

⇒2x²+x– 528 = 0

This is a Quadratic Equation.

(ii) Let two consecutive numbers be x and (x + 1).

It is given that x (x + 1) = 306

⇒ x²+x=306

⇒ x²+x-306=0

This is a Quadratic Equation.

(iii) Let present age of Rohan = 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:

(x + 3) (x + 29) = 360

⇒ x²+29x+3x+87=360

⇒ x²+32x-273=0

This is a Quadratic Equation.

(iv) Let speed of train be x km/h

Time taken by train to cover 480 km = 480 x hours

If speed had been 8km/h less then time taken would be (480x−8) hours

According to given condition,

if speed had been 8km/h less then time taken is 3 hours less.

Therefore, 480x – 8 = 480x + 3

⇒ 480 (1x – 8 − 1x) = 3

⇒ 480 (x – x + 8) (x) (x − 8) = 3

⇒ 480 × 8 = 3 (x) (x − 8)

⇒ 3840 = 3x²-24x

⇒ 3x²-24x-3840=0

Dividing equation by 3, we get

⇒x²-8x-1280=0

This is a Quadratic Equation

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