EXERCISE 4.4 CLASS 10TH MATHS NCERT SOLUTION QUESTION 2 CHAPTER 4 QUADRETIC EQUATION
प्रश्न -(2) निम्न प्रत्येक द्विघात समीकरण में K का ऐसा मान ज्ञात कीजिये की बराबर मूल हो।
( i ) 2x^2 + k x + 3 = 0 (ii ) k x( x - 2 ) + 6 = 0
हल : ( i ) दिया गया समीकरण , 2x^2 + k x + 3 = 0 में
a = 2 , b = k , c = 3
चूँकि मूल बराबर हे ,
इसलिए , b^2 - 4 *a *c = 0
=> k^2 - 4 *2 *3 = 0
=> k^2 - 24 = 0
=> k^2 = 24
=> k = \/24
=> k = +2\/6 , -2 \/6 : उत्तर
(ii ) दिया गया समीकरण, k x (x - 2 ) + 6 = 0
k x^2 - 2kx + 6 = 0
यहाँ a = k , b = -2k , c = 6
बराबर मुलो के लिए , b^2 - 4 *a *c = 0
=> (-2k )^ 2 - 4 * k * 6 = 0
=> 4k^2 -24k = 0
4 से भाग देने पर ,
=> k^2 - 6k = 0
=> k (k - 6 ) = 0
k = 0 तब 6 = 0 होगा जो संभव नहीं। k - 6 = 0 => k = 6 होगा।
ENGLISH TRANSLATION
Question -(2) In each of the following quadratic equations, find the value of K such that it has equal roots.
( i ) 2x^2 + k x + 3 = 0 (ii ) k x( x - 2 ) + 6 = 0
Solution : ( i ) In the given equation, 2x^2 + k x + 3 = 0
a = 2 , b = k , c = 3
Since the root is equal,
Therefore, b^2 - 4 *a *c = 0
=> k^2 - 4 *2 *3 = 0
=> K^2 - 24 = 0
=> k^2 = 24
=> k = \/24
=> k = +2\/6 , -2 \/6 : Answer
(ii) Given equation, k x (x - 2 ) + 6 = 0
k x^2 - 2kx + 6 = 0
Here a = k , b = -2k , c = 6
For equal roots, b^2 - 4 *a *c = 0
=> (-2k )^ 2 - 4 * k * 6 = 0
=> 4k^2 -24k = 0
On dividing by 4,
=> k^2 - 6k = 0
=> k (k - 6) = 0
k = 0 then 6 = 0 which is not possible. k - 6 = 0 => k = 6.