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## Karnataka State Syllabus Class 10 Maths Chapter 10 Quadratic Equations Ex 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}

Answer:

i) (x + 1)^{2} = 2(x – 3)

x^{2} + 2x + 1 = 2x – 6

x^{2} + 2x – 2x +1 + 6 = 0

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

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

∴ This is a quadratic equation.

ii) x^{2} – 2x = (-2) (3 – x)

x^{2} – 2x = -6 + 2

x^{2} – 2x – 2x + 6 = 0

x^{2} – 4x + 6 = 0

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

∴ This 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

-x – 2x – 2 + 3 = 0

-3x + 1 = 0

3x – 1 = 0

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

∴ Given equation is not quadratic equation.

iv) (x – 3) (2x + 1) = x (x + 5)

2x^{2} + x – 6x – 3 = x^{2} + 5x

2x^{2} – 5x – 3 = x^{2} + 5x

2x^{2} – x^{2} – 5x – 5x – 3 = 0

x^{2} – 10x – 3 = 0

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

∴ This 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

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

∴ This is a quadratic equation.

vi) x^{2} + 3x + 1 = {x – 2}^{2}

x^{2} + 3x + 1 = x^{2} – 4x + 4

x^{2} + 3x + 1 – x^{2} + 4x – 4 = 0

7x – 3 = 0

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

∴ Given equation is not quadratic equation.

vii) (x + 2)^{3} = 2x (x^{2} – 1)

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

x^{3} + 8 = 6x(x + 2) = 2x^{3} – 2x

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

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

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

x^{3} – 6x^{2} – 14x – 8 = 0

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

∴ Given equation is not quadratic equation.

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

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

x^{3} – 4x^{2} – x + 1 = x^{2} – 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

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

∴ This is a quadratic equation.

Question 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.

Answer:

Let the breadth of the rectangular Plot be ’x’ metres and the length of the rectangular Plot be (2x + 1)

Area of Rectangular Plot

= Length × breadth

528 = (2x + 1) x

528 = 2×2 + x

2x^{2} + x – 528 = 0

∴Now it is in the form of ax2 + bx + c = 0

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

Answer:

Let two Positive consecutive integers be x and x + 1

∴ Product = 306

x(x + 1) = 306

x^{2} + x = 306

x^{2} + x – 306 = 0

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

(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.

Answer:

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

3 years from now,

age of Rohan = (x + 3) and

age of Rohan mother = (x + 26 +3) = x + 29

Product of their ages = 360

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

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

x^{2} + 32x + 87 – 360 = 0

x^{2} + 32x – 273 = 0

(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.

Answer:

Let the uniform speed of the train be x km/hr

Time taken by the train to cover 480 km

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

If the speed of the train be 8 km/hr less, then the speed be (x – 8) km/hr

Time taken to cover distance 480 km

\(\frac{480}{x – 8}\)

difference in time.

3840 = 3x^{2} – 24x

3x^{2} – 24x – 3840 = 0 divide by 3

x^{2} – 8x – 1280 = 0