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.

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.

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.
Reprsent 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.
Solution:
Let the breadth of rectangular plot (b) be ’x’ m.
Then the length of th plot is one more than twice its breadth,
∴ Length (l)= 2x + 1 m.
But Length × Breadth = Area of rectangle
l × b = A
∴ x × (2x + 1) = 528 sq.m.
2x2 + x = 528
∴ 2x2 + x – 528 = 0 is the required equation.
Now, we have to find out the value of ‘x’ :
2x2 + x – 528 = 0
2x2 – 32x + 33x – 528 = 0
2x(x – 16) + 33(x – 16) = 0
(x – 16) (2x + 33) = 0
If x – 16 = 0, then x = 16
If 2x + 33 = 0, then x = -33/2
∴ Breadth (b) = 16 m.
Length (l) = (2x + 1) = 2(16) + 1 = 32 + 1 = 33m
∴ Length (l) = 33 m
Breadth (b) = 16 m.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Solution:
Let one positive integer be x’.
The Next integer is (x + 1)
Their product is 306.
∴ x (x + 1) = 306
x2 + x = 306
∴ x2 + x – 306= 0. This is required equation.
Now, we have to solve for positive integer.
x2 + x – 306 = 0
x2 + 18x – 17x – 306 = 0
x(x + 18) – 17(x + 18) = 0
(x + 18) (x – 17) = 0
If x + 18 = 0, then x = -18
If x – 17 = 0, then x = 17
∴ x = 18, OR x = 17.

(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.
Solution:
Let the present age of Rohan be ‘x’, then His mother’s age will be (x + 26)
After 3 years, Age of Rohan is (x + 3). After 3 years his mother’s age will be
= (x + 26 + 3)
= (x + 29)
Then product of their ages is 360.
∴ (x + 3) (x + 29) = 360
x2 + 29x + 3x + 87 = 360
x2 + 32x + 87 = 360
x2 + 32x + 87 – 360 = 0
x2 + 32x – 273 = 0.
This is the required equation.
Now, we have to solve for the value of ‘x’:
x2 + 32x – 273 = 0
x2 + 39x – 7x – 273 = 0
x(x + 39) – 7(x + 39) = 0
(x + 39) (x – 6) = 0
If x + 39 = 0, then x = -39
If x – 6 = 0, then x = 7
Present age of Rohan’s mother
= x + 26
= 7 + 26
= 33 years.

(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:
Let the initial speed of a train be ‘x’ km/h.
Time required to travel x km is 1 hour. Time required to travel 480 km ………?
\(\frac{480}{x}\) hr
If its speed decreases to 8 km/h, then it is (x – 8) km/h.
Time required to cover (x – 8) km is 1 Hr.
Time required to cover 480 km ………..?
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.1 1
∴ x(3x + 456) = 480 (x – 8)
3x2 + 456x = 480x + 3840
3x2 + 456x – 480 x + 3840 = 0
3x2 – 24x + 3840 = 0
∴ x2 – 8x + 1280 = 0
This is the required equation.
Now, we have to solve for x :
x2 – 8x + 1280 = 0
x2 – 40x + 32x + 1280 = 0
x(x – 40) + 32(x + 40) = 0
(x – 40) (x + 32) = 0
If x – 40 = 0, then x = 40
If x + 32 = 0, then x = -32
∴ Average speed of train is 40 km/hr.

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