Punjab State Board PSEB 10th Class Maths Book Solutions Chapter 4 Quadratic Equations Ex 4.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 10 Maths Chapter 4 Quadratic Equations Ex 4.1

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)

(vii) (x + 2)^{3} = 2x(x^{2} – 1)

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

Solution:

(i) Given that

(x + 1)^{2} = 2(x – 3)

Or x^{2} + 1 + 2x = 2x – 6

Or x^{2} + 1 + 2x – 2x + 6 = 0

Or x^{2} + 7 = 0

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

which is in the formof ax^{2} + bx + c = 0;

∴ It is a quadratic equation.

(ii) Given that

x^{2} – 2x = (-2) (3 – x)

Or x^{2} – 2x = -6 + 2x

Or x^{2} – 2x + 6 – 2x = 0

Or x^{2} – 4x + 6 = 0

which is the form of ax^{2} + bx + c = 0; a ≠ 0

∴ It is the quadratic equation.

(iii) Given that ,

(x – 2) (x + 1) = (x – 1) (x + 3)

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

Or x^{2} – x – 2 = x^{2} + 2x – 3

Or x^{2} – x – 2 – x^{2} -2x + 3 = 0

Or -3x + 1 = 0 which have no term of x^{2}.

So it is not a quadratic equation.

(iv) Given that

(x – 3)(2x + 1) = x(x + 5)

Or 2x^{2} + x – 6x – 3 = x^{2} + 5x

Or 2x^{2} – 5x – 3 – x^{2} – 5x = 0

Or x^{2} – 10x – 3 = 0

which is a form of ax^{2} + bx + c = 0; a ≠ 0

∴ It is a quadratic equation.

(v) Given that ,

(2x – 1) (x – 3) = (x + 5) (x – 1)

0r2x^{2} – 6x – x + 3 = x^{2} – x + 5x – 5

Or 2x^{2} – 7x + 3 = x^{2} + 4x – 5

Or 2x^{2} – 7x + 3 – x^{2} – 4x + 5 = 0

Or x^{2} – 11x + 8 = 0

which is a form of ax^{2} + bx + c = 0; a ≠ 0

∴ It is a quadratic equation.

(vi) Given that

x^{2}+3x+1 = (x – 2)^{2}

Or x^{2} + 3x + 1 = x^{2} + 4 – 4x

Or x^{2} + 3x + 1 – x^{2} – 4 + 4x = 0

Or 7x – 3 = 0

which have no term of x^{2}.

So it is not a quadratic equation.

(vii) Given that

(x + 2)^{3} = 2x(x^{2} – 1)

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

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

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

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

Here the highest degree of x is 3. which is a cubic equation.

∴ It is not a quadratic equation.

(viii) Given that

x^{3} – 4x^{2} – x+ 1= (x – 2)^{3}

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

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

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

Or 2x^{2} – 13x + 9 = 0

which is in the form of ax^{2} + bx +c = 0; a ≠ 0

∴ It 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.

(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) 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 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:

(i) Let Breadth of rectangular plot = x m

Length of rectangular plot= (2x + 1) m

∴ Area of rectangular plot = [x (2x + 1)] m^{2} = (2x^{2} + x) m^{2}

According to question,

2x^{2} + x = 528

S = 1

P = -528 × 2 = -1056

0r 2x^{2} + x – 528 = 0

Or 2x^{2} – 32x + 33x – 528 = 0

Or 2x(x – 16) + 33(x – 16) = 0

Or (x – 16) (2x + 33) = 0

Either x – 16 = 0 Or 2x + 33 = 0

x = 16 Or x = 2

∵ breadth of any rectangle cannot be negative, so we reject x = \(\frac{-33}{2}\), x = 16

Hence, breadth of rectangular plot = 16 m

Length of rectangular plot = (2 ×16 + 1)m = 33m

and given problem in the form of Quadratic Equation are 2x^{2} + x – 528 = 0.

(ii) Let two consecutive positive integers are x and x + 1.

Product of Integers = x (x + 1) = x^{2} + x

According to question,

Or x^{2} + x – 306 = 0

S = 1, P = – 306

Or x^{2} + 18x – 17x – 306 = 0

Or x(x + 18) -17 (x + 18) = 0

Or (x + 18) (x – 17) = 0

Either x + 18 = 0 Or x – 17 = 0

x = -18 Or x = 17

∵ We are to study about the positive integers, so we reject x = – 18.

x = 17

Hence, two consecutive positive integers are 17, 17 + 1 = 18

and given problem in the form of Quadratic Equation is x^{2} + x – 306 = 0.

(iii) Let present age of Rohan = x years

Rohan’s mother’s age = (x + 26) years

After 3 years, Rohan’s age = (x + 3) years

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

∴ Their product = (x + 3) (x + 29)

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

= x^{2} + 32x + 87

According to question,

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

Or x^{2} + 32x + 87 – 360 = 0

Or x^{2} + 32x – 273 = 0

Or x^{2} + 39x – 7x – 273 = 0

S = 32, P = – 273

Or x(x + 39) – 7(x + 39) = 0

Or (x + 39) (x – 7) =

Either x + 39 = Or x – 7 = 0

x = -39 Or x = 7

∵ age of any person cannot be negative so, we reject x = -39

∴ x = 7

Hence, Rohans present age = 7 years

and given problem in the form of Quadratic Equation is x^{2} + 32x – 273 = 0.

(iv) Let u km/hour be the speed of train.

Distance covered by train = 480 km

Time taken by train = \(\frac{480}{u}\) hour

[ Using, Speed = \(\frac{\text { Distance }}{\text { Time }}\)

or Time = \(=\frac{\text { Distance }}{\text { Speed }}\) ]

If speed of train be decreased 8km/hr.

∴ New speed of train = (u – 8) km/hr.

and time taken by train = \(\frac{480}{u-8}\) hour

According to question.

or 3840 = 3 (u^{2} – 8u)

or u^{2} – 8u = 1280

or u^{2} – 8u – 1280=0

or u^{2} – 40u + 32u – 1280 = 0

S = -8, P = – 1280

or u(u – 40) + 32 (u – 40) = 0

or (u – 40)(u + 32) = 0

Either u – 40 = 0

or u + 32 = 0

u = 40 or u = -32

But, speed cannot be negative so we reject

u = – 32

∴ u = 40.

Hence speed of train is 40 km/hr Ans.