PSEB 10th Class Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

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) x2 – 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) x2 + 3x + 1 = (x – 2)
(vii) (x + 2)3 = 2x(x2 – 1)
(viii) x3 – 4x2 – x + 1 = (x – 2)3

Solution:
(i) Given that
(x + 1)2 = 2(x – 3)
Or x2 + 1 + 2x = 2x – 6
Or x2 + 1 + 2x – 2x + 6 = 0
Or x2 + 7 = 0
Or x2 + 0x + 7 = 0
which is in the formof ax2 + bx + c = 0;
∴ It is a quadratic equation.

(ii) Given that
x2 – 2x = (-2) (3 – x)
Or x2 – 2x = -6 + 2x
Or x2 – 2x + 6 – 2x = 0
Or x2 – 4x + 6 = 0
which is the form of ax2 + bx + c = 0; a ≠ 0
∴ It is the quadratic equation.

(iii) Given that ,
(x – 2) (x + 1) = (x – 1) (x + 3)
Or x2 + x – 2x – 2 = x2 + 3x – x – 3
Or x2 – x – 2 = x2 + 2x – 3
Or x2 – x – 2 – x2 -2x + 3 = 0
Or -3x + 1 = 0 which have no term of x2.
So it is not a quadratic equation.

(iv) Given that
(x – 3)(2x + 1) = x(x + 5)
Or 2x2 + x – 6x – 3 = x2 + 5x
Or 2x2 – 5x – 3 – x2 – 5x = 0
Or x2 – 10x – 3 = 0
which is a form of ax2 + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.

(v) Given that ,
(2x – 1) (x – 3) = (x + 5) (x – 1)
0r2x2 – 6x – x + 3 = x2 – x + 5x – 5
Or 2x2 – 7x + 3 = x2 + 4x – 5
Or 2x2 – 7x + 3 – x2 – 4x + 5 = 0
Or x2 – 11x + 8 = 0
which is a form of ax2 + bx + c = 0; a ≠ 0
∴ It is a quadratic equation.

(vi) Given that
x2+3x+1 = (x – 2)2
Or x2 + 3x + 1 = x2 + 4 – 4x
Or x2 + 3x + 1 – x2 – 4 + 4x = 0
Or 7x – 3 = 0
which have no term of x2.
So it is not a quadratic equation.

(vii) Given that
(x + 2)3 = 2x(x2 – 1)
Or x3 + (2)3 + 3 (x)2 2 + 3(x)(2)2 = 2x3 – 2x
Or x3 + 8 + 6x2 + 12x = 2x3 – 2x
Or x3 + 8 + 6x2 + 12x – 2x3 + 2x = 0
Or -x3 + 6x2 + 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
x3 – 4x2 – x+ 1= (x – 2)3
Or x3 – 4x2 – x + 1 = x3 – (2)3 + 3(x)2 (-2) + 3 (x) (-2)2
Or x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x
Or x3 – 4x2 – x + 1 – x3 + 8 + 6x2 – 12x = 0
Or 2x2 – 13x + 9 = 0
which is in the form of ax2 + 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 m2. 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)] m2 = (2x2 + x) m2
According to question,
2x2 + x = 528
S = 1
P = -528 × 2 = -1056
0r 2x2 + x – 528 = 0
Or 2x2 – 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 2x2 + x – 528 = 0.

(ii) Let two consecutive positive integers are x and x + 1.
Product of Integers = x (x + 1) = x2 + x
According to question,
Or x2 + x – 306 = 0
S = 1, P = – 306
Or x2 + 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 x2 + 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)
= x2 + 29x + 3x + 87
= x2 + 32x + 87
According to question,
x2 + 32x + 87 = 360
Or x2 + 32x + 87 – 360 = 0
Or x2 + 32x – 273 = 0
Or x2 + 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 x2 + 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 (u2 – 8u)
or u2 – 8u = 1280
or u2 – 8u – 1280=0
or u2 – 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.