Students can Download Class 10 Maths Chapter 10 Quadratic Equations Ex 10.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.
Karnataka State Syllabus Class 10 Maths Chapter 10 Quadratic Equations Ex 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
Answer:
i) (x + 1)2 = 2(x – 3)
x2 + 2x + 1 = 2x – 6
x2 + 2x – 2x +1 + 6 = 0
x2 + 0x + 7 = 0
This is in the form of ax2 + bx + c = 0.
∴ This is a quadratic equation.
ii) x2 – 2x = (-2) (3 – x)
x2 – 2x = -6 + 2
x2 – 2x – 2x + 6 = 0
x2 – 4x + 6 = 0
This is in the form of ax2 + bx + c = 0.
∴ This is a quadratic equation.
iii) (x – 2) (x + 1) = (x – 1) (x + 3)
x2 + x – 2x – 2 = x2 + 3x – x – 3
x2 – x2 = x2 + 2x – 3
x2 – x2 – x – 2x – 2 + 3 = 0
-x – 2x – 2 + 3 = 0
-3x + 1 = 0
3x – 1 = 0
This is not in the form of ax2 + bx + c = 0
∴ Given equation is not 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
This is in the form of ax2 + bx + c = 0.
∴ This 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
This is in the form of ax2 + bx + c = 0.
∴ This is a quadratic equation.
vi) x2 + 3x + 1 = {x – 2}2
x2 + 3x + 1 = x2 – 4x + 4
x2 + 3x + 1 – x2 + 4x – 4 = 0
7x – 3 = 0
This is not in the form of ax2 + bx + c = 0
∴ Given equation is not quadratic equation.
vii) (x + 2)3 = 2x (x2 – 1)
x3 + 8 + 3(x)(2)(x+2) = 2x3 – 2x
x3 + 8 = 6x(x + 2) = 2x3 – 2x
x3 + 8 + 6x2 + 12x = 2x3 – 2x
x3 – 2x3 + 6x2 + 12x + 2x + 8 = 0
-x3 + 6x2 + 14x + 8 = 0
x3 – 6x2 – 14x – 8 = 0
This is not in the form of ax2 + bx + c = 0
∴ Given equation is not quadratic equation.
viii) x3 – 4x2 – x + 1 = (x – 2)3
x3 – 4x2 – x + 1 = x3 – 8 – 3(x)(2)(x – 2)
x3 – 4x2 – x + 1 = x2 – 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
This is in the form of ax2 + bx + c = 0.
∴ This 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.
Answer:
Let the breadth of the rectangular Plot be ’x’ metres and the length of the rectangular Plot be (2x + 1)
Area of Rectangular Plot
= Length × breadth
528 = (2x + 1) x
528 = 2×2 + x
2x2 + x – 528 = 0
∴Now it is in the form of ax2 + bx + c = 0
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Answer:
Let two Positive consecutive integers be x and x + 1
∴ Product = 306
x(x + 1) = 306
x2 + x = 306
x2 + x – 306 = 0
∴It is in the form ax2 + bx + c = 0
(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.
Answer:
Let Rohan’s present age be x years then, present age of Rohan’s mother be (x + 26) years
3 years from now,
age of Rohan = (x + 3) and
age of Rohan mother = (x + 26 +3) = x + 29
Product of their ages = 360
(x + 3) (x + 29) = 360
x2 + 29x + 3x + 87 = 360
x2 + 32x + 87 – 360 = 0
x2 + 32x – 273 = 0
(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.
Answer:
Let the uniform speed of the train be x km/hr
Time taken by the train to cover 480 km
\(\frac{480}{x}\)
If the speed of the train be 8 km/hr less, then the speed be (x – 8) km/hr
Time taken to cover distance 480 km
\(\frac{480}{x – 8}\)
difference in time.
3840 = 3x2 – 24x
3x2 – 24x – 3840 = 0 divide by 3
x2 – 8x – 1280 = 0