# Class10 NCRT Quadratic Equations Exercise – 4.1 pdf || UP Board

# __Quadratic Equations__

__Exercise – 4.1__

__Quadratic Equations__

__Exercise – 4.1__

**1. Check whether the following are
quadratic equations****:**

**(****i****)**** (x + 1) ^{2} = 2(x – 3)**

on simplifying the equation,

(x + 1)^{2} = 2(x – 3)

x^{2} + 2x + 1 = 2x – 6

x^{2} + 7 = 0

Or x^{2} + 0x + 7 = 0

This equation is of the type ax^{2}
+ bx + c = 0.

Hence, the given equation is a
quadratic equation.

(ii) x^{2}
– 2x = (-2)(3 – x)

on simplifying the equation,

x^{2} – 2x = (-2)(3 - x)

x^{2} – 2x = -6 + 2x

x^{2} – 4x + 6 = 0

Or x^{2} – 4x + 6 = 0

This equation is of the type ax^{2}
+ bx + c = 0.

Hence, the given equation is a
quadratic equation.

(iii) (x –
2)(x + 1) = (x – 1)(x + 3)

on simplifying the equation,

(x – 2)(x + 1) = (x – 1)(x + 3)

x^{2} – 2x + x – 2 = x^{2}
– x + 3x – 3

-3x + 1 = 0

Or 3x – 1 = 0

This equation is not of the type ax^{2}
+ bx + c = 0.

Hence, the given equation is not a
quadratic equation.

(iv) (x –
3)(2x + 1) = x(x + 5)

on simplifying the equation,

2x^{2} – 6x + x – 3 = x^{2}
+ 5x

x^{2} – 10x – 3 = 0

Or x^{2} +(-10)x + (-3) = 0

This equation is of the type ax^{2}
+ bx + c = 0.

Hence, the given equation is a
quadratic equation.

(v) (2x –
1)(x – 3) = (x + 5)(x – 1)

on simplifying the equation,

2x^{2} – x – 6x + 3 = x^{2}
+ 5x – x – 5

x^{2} – 11x + 8 = 0

This equation is of the type ax^{2}
+ bx + c = 0.

Hence, the given equation is a
quadratic equation.

(vi) x^{2}
+ 3x + 1 = (x – 2)^{2}

on simplifying the equation,

x^{2} + 3x + 1 = x^{2}
– 4x + 4

7x – 3 =0

This equation is not of the type ax^{2}
+ bx + c = 0.

Hence, the given equation is not a
quadratic equation.

(vii) (x + 2)^{3}
= 2x(x^{2} – 1)

on simplifying the equation,

x^{3} + 6x^{2} + 12x
+ 8 = 2x^{3} – 2x

-x^{3} + 6x^{2} +
14x + 8 = 0

अर्थात x^{3} – 6x^{2} – 14x
– 8 = 0

This equation is not of the type ax2 + bx + c = 0.

Hence, the given equation is not a
quadratic equation.

(viii) x^{3}
– 4x^{2} – x + 1 = (x – 2)^{3}

on simplifying the equation,

x^{3} – 4x^{2} – x +
1 = x^{3} – 6x^{2} + 12x – 8

2x^{2} – 13x + 9 = 0

2x^{2} + (– 13)x + 9 = 0

This equation is of the type ax^{2}
+ bx + c = 0.

Hence, the given equation is a
quadratic equation.

**2. ****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 = x m

Hence, length of plot = 2x + 1 m

Hence, area = x(2x + 1) m^{2}

According to Question,

x(2x + 1) = 528

2x^{2} + x = 528

2x^{2} + x – 528 = 0

Hence, the quadratic equation 2x^{2}
+ x – 528 = 0 represents the length and breadth of the plot.

(ii) The product of two consecutive
positive integers is 306. we need to find the integers.

Let first integer = x

Therefore, second integer = x + 1

So, product = x(x + 1)

According to Question,

x(x + 1) = 306

x^{2} + x = 306

x^{2} + x – 306 = 0

Hence, the quadratic equation x^{2}
+ x – 306 = 0 represents the product of two consecutive positive integers.

(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 Rohon’s present age.

Let Rohan's age = x years

Therefore, age of Rohan's mother = x
+ 26 years

after three years,

Let Rohan's age = x + 3 years

Therefore, age of Rohan's mother = x
+ 29 years

Therefore, product of ages = (x + 3)(x + 29) years

According to Question,

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

x^{2} + 3x + 29x + 87 = 360

x^{2} + 32x – 273 = 0

Hence, the quadratic equation x^{2}
+ 32x – 273 = 0 represents the Rohon’s present age.

(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.

Let speed of train = x km/h

Total distance = 480 km

Hence, time taken by the train = 480/x hours

If its speed had been 8 km/m less, then time taken = 480/(x – 8) hours

According to Question,

480x – 480x + 3640 = 3(x – 8)x

3640 = 3x^{2} – 24x

3x^{2} – 24x – 3640 = 0

Hence, the
quadratic equation 3x^{2} – 24x – 3640 = 0 represents the speed of the
train.

Download Class10 NCRT Quadratic Equations Exercise – 4.1 pdf

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