AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.1 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.1

### 10th Class Maths 5th Lesson Quadratic Equations Ex 5.1 Textbook Questions and Answers

Question 1.

Check whether the following are quadratic equations.

i) (x + l)^{2} = 2(x-3)

Answer:

Given: (x + l)^{2} = 2(x – 3)

⇒ x^{2} + 2x + 1 = 2(x – 3) = 2x – 6

⇒ x^{2} + 2x + l – 2x + 6 = 0

⇒ x^{2} + 7 = 0 is a Q.E.

ii) x^{2} – 2x = (-2) (3 – x)

Answer:

Given: x^{2} – 2x = -2(3 – x)

⇒ x^{2} – 2x = -6 + 2x

⇒ x^{2} – 4x + 6 = 0 is a Q.E.

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

Answer:

Given: (x – 2) (x + 1) = (x – 1) (x + 3)

⇒ x (x + 1) – 2 (x +1)

= x (x + 3) – 1 (x + 3)

Note : Compare the coefficients of x2 on both sides. If they are equal it is not a Q.E.

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

⇒ x^{2} – x – 2 = x^{2} + 2x – 3

⇒ 3x – 1 = 0 is not a Q.E.

iv) (x – 3) (2x + 1) = x(x + 5)

Answer:

Given: (x – 3) (2x + 1) = x(x + 5)

⇒ x (2x + 1) – 3 (2x + 1) = x . x + 5 . x

⇒ 2x^{2} + x – 6x – 3 = x^{2} + 5x

⇒ 2x^{2} – 5x – 3 – x^{2} – 5x = 0

⇒ x^{2} – 10x – 3 = 0 is a Q.E.

(or)

Comparing the coefficients of x2 on both sides.

x . 2x and x . x

⇒ 2x^{2} and x^{2}

2x^{2} ≠ x^{2}

Hence it’s a Q.E.

v) (2x – 1) (x – 3) = (x + 5) (x – 1)

Answer:

Given: (2x – 1) (x – 3) = (x + 5) (x – 1)

⇒ 2x (x – 3) -1 (x – 3) = x (x – 1) + 5(x – 1)

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

⇒ 2x^{2} -7x + 3 – x^{2} – 4x + 5 = 0

⇒ x^{2} – 11x + 8 = 0

Hence it’s a Q.E.

(or)

Co.eff. of x^{2} on L.H.S. = 2 × 1 = 2

Co.eff. of x^{2} on R.H.S = 1 × 1 = 1

LHS ≠ RHS Hence it is a Q.E.

vi) x^{2} + 3x + 1 = (x – 2)^{2}

Answer:

Given: x^{2} + 3x + 1 = (x – 2)^{2}

⇒ x^{2} + 3x + 1 = x^{2} – 4x + 4

⇒ 7x – 3 = 0 is not a Q.E.

vii) (x + 2)^{3} = 2x (x^{2} – 1)

Answer:

Given: (x + 2)^{3} = 2x(x^{2} – 1)

⇒ x^{3} + 6x^{2} + 12x + 8 = 2x^{3} – 2x [∵ (a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}]

⇒ -x^{3} + 6x^{2} + 14x + 8 = 0

is not a Q.E. [∵ degree = 3]

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

Answer:

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

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

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

⇒ 2x^{2} – 13x + 9 = 0 is a Q.E.

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 meters) 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 m.

Then its length (by problem) = 2x + 1.

Area = l . b = (2x + 1) . x = 2x^{2} + x

But area = 528 m^{2} (∵ given)

∴ 2x^{2} + x = 528

⇒ 2x^{2} + x – 528 = 0 where x is the breadth of the rectangle.

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

Answer:

Let the consecutive integers be x and x + 1.

Their product = x(x + 1) = x^{2} + x

By problem x^{2} + x = 306

⇒ x^{2} + x – 306 = 0

where x is the smaller integer.

iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.

Answer:

Let the present age of Rohan be x years.

Then age of Rohan’s mother = x + 26

After 3 years:

Age of Rohan would be = x + 3

Rohan’s mother’s age would be = (x + 26) + 3 = x + 29

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

⇒ x(x + 29) + 3(x + 29) = 360

⇒ x^{2} + 29x + 3x + 87 = 360

⇒ x^{2} + 32x + 87 – 360 = 0

⇒ x^{2} + 32x – 273 = 0

⇒ x^{2} + 39x – 7x – 273 = 0

⇒ x (x + 39) – 7 (x + 39) = 0

⇒ (x – 7) (x + 39) = 0

⇒ x = 7 or x = -39 ‘x’ being age cannot be negative.

∴ x = Present age of Rohan = 7 years.

iv) A train travels a distance of 480 km at a 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 the speed of the train.

Answer:

Let the speed of the train be x km/h.

Then time taken to travel a distance of distance of 480 km = \(\frac{\text { distance }}{\text { speed }}\) = \(\frac{480}{x}\)

If the speed is 8km/h less, then time needed to cover the same distance would be \(\frac{480}{x-8}\)

⇒ x^{2} – 8x = 1280

⇒ x^{2} – 8x – 1280 = 0

where x is the speed of the train.