KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1

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) 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)2
vii) (x + 2)3 = 2x (x2 – 1)
viii) x3 – 4x2 – x + 1 = (x – 2)3
Solution:
i) (x + 1)2 = 2(x – 3)
x2 + 2x + 1 = 2x – 6
x2 + 2x – 2x + 1 – 6 = 0
x2 – 5 = 0
It is in the form of ax2 + c = 0
Therefore, the given equation is a Quadratic equation.

ii) x2 – 2x = (- 2) (3 – x)
x2 – 2x = – 6 + 2x
x2 – 2x – 2x + 6 = 0
x2 – 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)
x2 + x – 2x – 2 = x2 + 3x – x – 3
x2 – x – 2 = x2 + 2x – 3
x2 – x2 – x – 2x – 2 + 3 = 0
– 3x + 1 = 0
It is in not in the form of ax2 + bx + c = 0
Therefore, the given equation is not a Quadratic equation.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

iv) (x – 3) (2x + 1) = x (x + 5)
2x2 + x – 6x – 3 = x2 + 5x
2x2 – 5x – 3 = x2 + 5x
2x2 – x2 – 5x – 5x – 3 = 0
x2 – 10x – 3 = 0
It is in the form of ax2 + bx + c = 0
Therefore, the- given equation is a Quadratic equation.

v) (2x – 1) (x – 3) = (x + 5) (x – 1)
2x2 – 6x – x + 3 = x2 – x + 5x – 5
2x2 – 7x + 3 = x2 + 4x – 5
2x2 – x2 – 7x – 4x + 3 + 5 = 0
x2 – 11x + 8 = 0
It is in the form of ax2 + bx + c = 0
Therefore, the given equation is a Quadratic equation.

vi) x2 + 3x + 1 = (x – 2)2
x2 + 3x + 1 = x2 – 4x + 4
x2 – x2 + 3x + 4x + 1 – 4 = 0
7x – 3 = 0
It is not in the form of ax2 + bx + c = 0
Therefore, the given equation is not a Quadratic equation.

vii) (x + 2)3 = 2x (x2 – 1)
x3 + (3)3 + 3 × 2 × x (x + 2) = 2x3 – 2x
x3 + 27 + 6x (x + 2) = 2x3 – 2x
x3 + 27 + 6x2 + 12x = 2x3 – 2x
2x3 – x3 – 6x2 – 2x – 12x – 27 = 0
x3 – 6x2 – 14x – 27 = 0
It is not in the form of ax2 + bx + c = 0
Therefore, the given equation is not a Quadratic equation.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

viii) x3 – 4x2 – x + 1 = (x – 2)3
x3 – 4x2 – x + 1 = x3 – (2)3 – 3 × 2 × x(x – 2)
x3 – 4x3 – x + 1 = x3 – 8 – 6x (x – 2)
x3 – 4x2 – x + 1 = x3 – 8 – 6x2 + 12x
x3 – x3 – 4x2 + 6x2 – x – 12x + 1 + 8 = 0
2x2 – 13x + 9 = 0
It is in the form of ax2 + bx + c = 0
Therefore, the given equation 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/hr 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 the breadth = x metres
∵ Length = 2(Breadth) + 1
∴ Length = (2x + 1)metres
Since, Length × Breadth = Area
∴ (2x + 1) × x = 528 ⇒ 2x2 + x = 528
⇒ 2x2 + x – 528 = 0
Thus, the required quadratic equation is 2x2 + x – 528 = 0

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

(ii) Let the two consecutive positive integers be x and (x + 1).
∵ Product of two consecutive positive integers = 306
∴ x(x + 1) = 306 ⇒ x2 + x = 306
⇒ x2 + x – 306 = 0
Thus, the required quadratic equation is x2 + x – 306 = 0

(iii) Let the present age of Rohan be x years
∴ Mother’s age = (x + 26) years
After 3 years,
Rohan’s age = (x + 3) years
Mother’s age = [(x + 26) + 3] years
= (x + 29) years
According to the condition,
(x + 3) × (x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 29x + 3x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Thus, the required quadratic equation is x2 + 32x – 273 = 0

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

(iv) Let the speed of the train = u km/hr Distance covered = 480 km Distance
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1 1
In second case,
Speed = (u – 8) km/hour
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1 2
⇒ 480u – 480(u – 8) = 3u(u – 8)
⇒ 480u – 480u + 3840 = 3u2 – 24u
⇒ 3840 – 3u2 + 24u = 0
⇒ u2 – 8u – 1280 = 0
Thus, the required quadratic equation is u2 – 8u – 1280 = 0

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.1, drop a comment below and we will get back to you at the earliest.

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