NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations द्विघात समीकरण
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Exercise 4.1
Question 1:
Check whether the following are quadratic equations :
जाँच कीजिए कि क्या निम्न द्विघात समीकरण है:
(i) (x + 1)2 = 2 (x – 3) \(\quad\quad\quad\quad\quad\) (ii) x2 – 2x = (-2) (3 – x)
(iii) (x – 2) (x + 1) = (x – 1)(x + 3) \(\quad\;\) (iv) (x – 3)(2x + 1) = x (x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1) \(\quad\) (vi) x2 + 3x + 1 = (x – 2)2
(vii) (x + 2)3 = 2x(x2 – 1) \(\quad\quad\quad\;\) (viii) x3 – 4x2 – x + 1 = (x – 2)3
Solution
(i) (x + 1)2 = 2 (x – 3)
\(\quad\) x2 + 1 + 2x = 2x – 6
\(\Rightarrow\) x2 + 1 + 2x – 2x + 6 = 0
\(\Rightarrow\) x2 + 7 = 0
Hence, it is quadratic equation.
अतः यह द्विघात समीकरण है।
(ii) x2 – 2x = (-2) (3 – x)
\(\quad\) x2 – 2x = -6 + 2x
\(\Rightarrow\) x2 – 2x + 6 – 2x = 0
\(\Rightarrow\) x2 – 4x + 6 = 0
Hence, it is quadratic equation.
अतः यह द्विघात समीकरण है।
(iii) (x – 2) (x + 1) = (x – 1)(x + 3)
\(\quad\) (x – 2) (x + 1) = (x – 1)(x + 3)
\(\Rightarrow\) x2 – x – 2 = x2 + 2x – 3
\(\Rightarrow\) x2 – x – 2 – x2 -2x + 3 = 0
\(\Rightarrow\) – 3x + 1 = 0
Hence, it is not quadratic equation.
अतः यह द्विघात समीकरण नहीं है।
(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
Hence, it is quadratic equation.
अतः यह द्विघात समीकरण है।
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
⇒ (2x – 1) (2x – 6 ) = (x + 5) (x – 1)
⇒ 4x2 – 12x – 2x + 6 = x2 + 4x – 5
⇒ 4x2 – 14x + 6 = x2 – x + 4x – 5
⇒ 4x2 – x2 – 14x – 4x + 6 + 5 = 0
⇒ 3x2 – 18x + 11 = 0
Hence, it is quadratic equation.
अतः यह द्विघात समीकरण है।
(vi) x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = x2 – 2x +4
⇒x2 – x2 + 4x + 3x + 1 – 4 = 0
⇒ 7x – 3 = 0
Hence, it is not quadratic equation.
अतः यह द्विघात समीकरण नहीं है।
(vii) (x + 2)3 = 2x(x2 – 1)
⇒ x3 + 8 + 6+ 12x = 2x3 – 2x
⇒ 2x3 – x3 – 6-12x + 2x – 8 = 0
⇒ x3 – 6x2 -10x – 8 =0
Hence, it is not quadratic equation.
अतः यह द्विघात समीकरण नहीं है।
(viii) x3 – 4x2 – x + 1 = (x – 2)3
x3 – 4x2– x + 1 = (x – 2 )3
⇒x3 – 4x2– x + 1 = x3 – 8 + 6x2 + 12x
⇒ x3 – x3 – 4x2+ 6x2 -12x + 1 = 0
⇒ 2x2 -13x + 1 = 0
Hence, it is quadratic equation.
अतः यह द्विघात समीकरण है।
Question 2:
Represent the following situations in the form of quadratic equations :
निम्न स्थितियों को द्विघात समीकरणों के रूप में निरुपित कीजिए :
(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.
(i) एक आयताकार भूखंड का क्षेत्रफल 528 मी2 है | क्षेत्र की लंबाई (मीटरों में) चौड़ाई के दुगुने से एक अधिक है | हमें भूखंड की लंबाई और चौड़ाई ज्ञात करनी है |
Solution
Let the breadth of the rectangular plot = x m
Hence, the length of the plot is (2x + 1) m.
Formula of area of rectangle = length × breadth = 528 m2
Putting the value of length and width, we get
(2x + 1) × x = 528
⇒ 2x2 + x =528
⇒ 2x2 + x – 528 = 0
एक आयताकार भूखंड का क्षेत्रफल = 528 m2
माना आयताकार भूखंड की चौड़ाई = x m
आयताकार भूखंड की लंबाई = 2x + 1 m
आयताकार भूखंड का क्षेत्रफल = 528 m2
लंबाई × चौड़ाई = 528
(2x + 1)x = 528
⇒ 2x2 + x = 528
⇒ 2x2 + x – 528 = 0
\(\quad\)
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(ii) दो क्रमागत धनात्मक पूर्णाकों का गुणनफल 306 है | हमें पूर्णाकों को ज्ञात करना है |
Solution
Let the first integer number = x
Next consecutive positive integer will = x + 1
Product of both integers = x × (x +1) = 306
⇒ x2 + x = 306
⇒ x2 + x – 306 = 0
माना पहला धनात्मक पूर्णाक = x
दूसरा धनात्मक पूर्णाक = x + 1
दो क्रमागत धनात्मक पूर्णाकों का गुणनफल = 306
पहला धनात्मक पूर्णाक × दूसरा धनात्मक पूर्णाक = 306
(x + 1)x = 306
⇒ x2 + x = 306
⇒ x2 + x – 306 = 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.
(iii) रोहन की माँ उससे 26 वर्ष बड़ी है |उनकी आयु (वर्षों में) का गुणनफल अब से तीन वर्ष पश्चात् 360 हो जाएगी| हमें रोहन की वर्तमान आयु ज्ञात करणी है |
Solution
Let take Rohan’s age = x years
Hence, his mother’s age = x + 26
3 years from now
Rohan’s age = x + 3
Age of Rohan’s mother will = x + 26 + 3 = x + 29
The product of their ages 3 years from now will be 360 so that
(x + 3)(x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 32x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
माना रोहन की वर्तमान आयु = x वर्ष
रोहन की माँ की आयु = (x + 26) वर्ष
तीन वर्ष पश्चात रोहन की आयु = (x + 3) वर्ष
तीन वर्ष पश्चात रोहन की माँ की आयु = (x + 26 + 3) = (x + 29) वर्ष
दोनो की आयु का गुणनफल = 306
(x + 29)(x + 3) = 306
⇒ x2 + 29x + 3x + 87 = 306
⇒ x2 + 32x + 87 = 306
⇒ x2 + 32x = 273
⇒ x2 + 32x – 273 = 0
चूँकि यह ax2 + bx + c = 0 के रूप में है| अत: यह द्विघात समीकरण है |
(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.
(iv) एक रेलगाड़ी 480 कीमी की दुरी समान चाल से तय करती है | यदि इसकी चाल 8 कीमी/घंटा कम होती, तो वह उसी दूरी को तय करने में 3 घंटे अधिक लेती | हमें रेलगाड़ी की चाल ज्ञात करनी है|
Solution
Let the speed of the train be x km/h. माना रेलगाड़ी की समान्य चाल x km/hr है |
Distance covered = 480 km तय की गयी दुरी = 480 km
Time, t = \(\frac{Distance}{Speed}\) = \(\frac{480}{x}\)
ATQ,
If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance.
\(\quad\) t + 3 = \(\frac{480}{x – 8}\)
\(\Rightarrow\) \(\frac{480}{x}\) + 3 = \(\frac{480}{x-8}\)
\(\Rightarrow\) 3 = \(\frac{480}{x – 8}\) – \(\frac{480}{x}\)
\(\Rightarrow\) 480 (\(\frac{1}{x – 8}\) – \(\frac{1}{x}\)) = 3
\(\Rightarrow\) \(\frac{x – x + 8}{(x – 8)x}\) = \(\frac{3}{480}\)
\(\Rightarrow\) \(\frac{8}{x^2 – 8x}\) = \(\frac{1}{160}\)
\(\Rightarrow x^2 – 8x = 160\times8\)
\(\Rightarrow x^2 – 8x – 1280 = 0\)
⇒ x2 – 8x – 1280 = 0
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations
Exercise 4.2
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations
Exercise 4.3
NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations
