**KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1.

## Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1

Question 1.

Check whether the following are quadratic equations:

i) (x + 1)^{2} = 2 (x – 3)

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

vii) (x + 2)^{3} = 2x (x^{2} – 1)

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

Solution:

i) (x + 1)^{2} = 2(x – 3)

x^{2} + 2x + 1 = 2x – 6

x^{2} + 2x – 2x + 1 – 6 = 0

x^{2} – 5 = 0

It is in the form of ax^{2} + c = 0

Therefore, the given equation is a Quadratic equation.

ii) x^{2} – 2x = (- 2) (3 – x)

x^{2} – 2x = – 6 + 2x

x^{2} – 2x – 2x + 6 = 0

x^{2} – 4x + 6 = 0

It is in the form of ax2 + bx + c = 0

Therefore, the given equation is a Quadratic equation.

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

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

x^{2} – x – 2 = x^{2} + 2x – 3

x^{2} – x^{2} – x – 2x – 2 + 3 = 0

– 3x + 1 = 0

It is in not in the form of ax^{2} + bx + c = 0

Therefore, the given equation is not a Quadratic equation.

iv) (x – 3) (2x + 1) = x (x + 5)

2x^{2} + x – 6x – 3 = x2 + 5x

2x^{2} – 5x – 3 = x^{2} + 5x

2x^{2} – x^{2} – 5x – 5x – 3 = 0

x^{2} – 10x – 3 = 0

It is in the form of ax^{2} + bx + c = 0

Therefore, the- given equation is a Quadratic equation.

v) (2x – 1) (x – 3) = (x + 5) (x – 1)

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

2x^{2} – 7x + 3 = x^{2} + 4x – 5

2x^{2} – x^{2} – 7x – 4x + 3 + 5 = 0

x^{2} – 11x + 8 = 0

It is in the form of ax^{2} + bx + c = 0

Therefore, the given equation is a Quadratic equation.

vi) x^{2} + 3x + 1 = (x – 2)^{2}

x^{2} + 3x + 1 = x^{2} – 4x + 4

x^{2} – x^{2} + 3x + 4x + 1 – 4 = 0

7x – 3 = 0

It is not in the form of ax^{2} + bx + c = 0

Therefore, the given equation is not a Quadratic equation.

vii) (x + 2)^{3} = 2x (x^{2} – 1)

x^{3} + (3)^{3} + 3 × 2 × x (x + 2) = 2x^{3} – 2x

x^{3} + 27 + 6x (x + 2) = 2x^{3} – 2x

x^{3} + 27 + 6x^{2} + 12x = 2x^{3} – 2x

2x^{3} – x^{3} – 6x^{2} – 2x – 12x – 27 = 0

x^{3} – 6x^{2} – 14x – 27 = 0

It is not in the form of ax^{2} + bx + c = 0

Therefore, the given equation is not a Quadratic equation.

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

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

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

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

x^{3} – x^{3} – 4x^{2} + 6x^{2} – x – 12x + 1 + 8 = 0

2x^{2} – 13x + 9 = 0

It is in the form of ax^{2} + bx + c = 0

Therefore, the given equation is a Quadratic equation.

Question 2.

Reprsent 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 rectangular plot (b) be ’x’ m.

Then the length of th plot is one more than twice its breadth,

∴ Length (l)= 2x + 1 m.

But Length × Breadth = Area of rectangle

l × b = A

∴ x × (2x + 1) = 528 sq.m.

2x^{2} + x = 528

∴ 2x^{2} + x – 528 = 0 is the required equation.

Now, we have to find out the value of ‘x’ :

2x^{2} + x – 528 = 0

2x^{2} – 32x + 33x – 528 = 0

2x(x – 16) + 33(x – 16) = 0

(x – 16) (2x + 33) = 0

If x – 16 = 0, then x = 16

If 2x + 33 = 0, then x = -33/2

∴ Breadth (b) = 16 m.

Length (l) = (2x + 1) = 2(16) + 1 = 32 + 1 = 33m

∴ Length (l) = 33 m

Breadth (b) = 16 m.

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

Solution:

Let one positive integer be x’.

The Next integer is (x + 1)

Their product is 306.

∴ x (x + 1) = 306

x^{2} + x = 306

∴ x^{2} + x – 306= 0. This is required equation.

Now, we have to solve for positive integer.

x^{2} + x – 306 = 0

x^{2} + 18x – 17x – 306 = 0

x(x + 18) – 17(x + 18) = 0

(x + 18) (x – 17) = 0

If x + 18 = 0, then x = -18

If x – 17 = 0, then x = 17

∴ x = 18, OR x = 17.

(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 the present age of Rohan be ‘x’, then His mother’s age will be (x + 26)

After 3 years, Age of Rohan is (x + 3). After 3 years his mother’s age will be

= (x + 26 + 3)

= (x + 29)

Then product of their ages is 360.

∴ (x + 3) (x + 29) = 360

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

x^{2} + 32x + 87 = 360

x^{2} + 32x + 87 – 360 = 0

x^{2} + 32x – 273 = 0.

This is the required equation.

Now, we have to solve for the value of ‘x’:

x^{2} + 32x – 273 = 0

x^{2} + 39x – 7x – 273 = 0

x(x + 39) – 7(x + 39) = 0

(x + 39) (x – 6) = 0

If x + 39 = 0, then x = -39

If x – 6 = 0, then x = 7

Present age of Rohan’s mother

= x + 26

= 7 + 26

= 33 years.

(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 the initial speed of a train be ‘x’ km/h.

Time required to travel x km is 1 hour. Time required to travel 480 km ………?

\(\frac{480}{x}\) hr

If its speed decreases to 8 km/h, then it is (x – 8) km/h.

Time required to cover (x – 8) km is 1 Hr.

Time required to cover 480 km ………..?

∴ x(3x + 456) = 480 (x – 8)

3x^{2} + 456x = 480x + 3840

3x^{2} + 456x – 480 x + 3840 = 0

3x^{2} – 24x + 3840 = 0

∴ x^{2} – 8x + 1280 = 0

This is the required equation.

Now, we have to solve for x :

x^{2} – 8x + 1280 = 0

x^{2} – 40x + 32x + 1280 = 0

x(x – 40) + 32(x + 40) = 0

(x – 40) (x + 32) = 0

If x – 40 = 0, then x = 40

If x + 32 = 0, then x = -32

∴ Average speed of train is 40 km/hr.

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