EXERCISE 4.4 CLASS 10TH QUESDTION (1) MATHS NCERT SOLUTIONS CHAPTER 4
EXERCISE 4.4 MATHS CLASS 10 IN HINDI
प्रश्न -(1) निम्न द्विघात समीकरणों के मुलो की प्रकृति ज्ञात कीजिये। यदि मुलो का अस्तित्व हो ,तो उन्हें ज्ञात कीजिये ?
( i ) 2x^2 - 3x + 5 = 0
हल : चूँकि 2x^2 - 3x + 5 = 0 में a = 2 , b = -3 , c = 5
तब विविक्तकर b^2 - 4ac = (-3)^2 - 4 * 2 * 5
= 9 - 40
= -31 < 0
यह स्थिति दर्शाती हे की वास्तविक मुलो का अस्तित्व नहीं हे।
(ii ) 3x^2 - 4\/3 x + 4 =0
हल : यहाँ a = 3 , b = -4 \/3 , c = 4
अब विविक्तकर b^2 - 4ac = (-4 \/3 )^2 - 4 * 3 * 4
= 48 - 48
b^2 - 4ac = 0 ------->(1 )
इस स्थिति में समीकरण के मूल वास्तविक एवं बराबर हे।
अब , (+) चिन्ह लेने पर समीकरण के मूल हे।
x = - b + \/ D / 2a
x = -b + \/b^2 - 4ac / 2a
x ={ - (-4 \/3 ) + \/0 / 2 * 3
x = 4\/3 + 0 / 6
x = 4 \/3 / 6
x = 2 / \/3
अब , (-) चिन्ह लेने पर समीकरण के मूल हे।
x = - b - \/ D / 2a
x = -b - \/b^2 - 4ac / 2a
x ={ - (-4 \/3 ) - \/0 / 2 * 3
x = 4\/3 - 0 / 6
x = 4 \/3 / 6
x = 2 / \/3
अतः समीकरण के अभीष्ट मूल x = 2 / \/3 , 2 / \/3 होंगे। :उत्तर
( नोट ; ^ का मतलब घात से हे )
(iii) चूँकि 2x^2 - 6x + 3 = 0
यहाँ a = 2 , b = -6 , c = 3
विविक्तकर => b^2 - 4ac = (-6 )^2 - 4 * 2 * 3
= 36 - 24
= 12 > 0 मूल असमान व वास्तविक हे।
मूल = -b +{ \/b^2 - 4 *a *c } / 2a (+ चिन्ह लेने पर )
= - (-6 ) + 12 / 2 * 2
= (6 +\/12) / 4
= (6 + 2 \/3 ) / 4
= 2 (3 + \/3 ) / 4
= 3 + \/3 / 2
इसी प्रकार , (-) चिन्ह लेने पर ,
मूल = -b - {\/b^2 - 4 a c } / 2a
= - (-6 ) - \/12 / 2 * 2
= (6 - 2 \/3 ) / 4
= 2 (3 - \/3 ) / 4
= 3 - \/3 / 2
उत्तर ; 3 + \/3 / 2 और 3 - \/3 / 2
ENGLISH TRANSLATION :
Question - (1) Find the nature of the roots of the following quadratic equations. If mullahs exist, find them?
( i ) 2x^2 - 3x + 5 = 0
Solution : Since 2x^2 - 3x + 5 = 0 in a = 2 , b = -3 , c = 5
Then the discriminant b^2 - 4ac = (-3)^2 - 4 * 2 * 5
= 9 - 40
= -31 < 0
This situation shows that real people do not exist.
(ii ) 3x^2 - 4\/3 x + 4 =0
Solution : Here a = 3 , b = -4 \/3 , c = 4
Now discriminant b^2 - 4ac = (-4 \/3 )^2 - 4 * 3 * 4
= 48 - 48
b^2 - 4ac = 0 ------->(1 )
In this case the roots of the equation are real and equal.
Now, taking the (+) sign, are the roots of the equation.
x = - b + \/ d / 2a
x = -b + \/b^2 - 4ac / 2a
x ={ - (-4 \/3 ) + \/0 / 2 * 3
x = 4\/3 + 0 / 6
x = 4 \/3 / 6
x = 2 / \/3
Now, taking the (-) sign, are the roots of the equation.
x = - b - \/ d / 2a
x = -b - \/b^2 - 4ac / 2a
x ={ - (-4 \/3 ) - \/0 / 2 * 3
x = 4\/3 - 0 / 6
x = 4 \/3 / 6
x = 2 / \/3
Hence, the required roots of the equation will be x = 2 / \/3 , 2 / \/3. :answer
(iii) Since 2x^2 - 6x + 3 = 0
Here a = 2 , b = -6 , c = 3
Discount => b^2 - 4ac = (-6 )^2 - 4 * 2 * 3
= 36 - 24
= 12 > 0 The root is unequal and real.
Root = -b +{ \/b^2 - 4 *a *c } / 2a (taking the + sign)
= - (-6 ) + 12 / 2 * 2
= (6 +\/12) / 4
= (6 + 2 \/3 ) / 4
= 2 (3 + \/3 ) / 4
= 3 + \/3 / 2
Similarly, taking the (-) sign,
Root = -b - {\/b^2 - 4 a c } / 2a
= - (-6 ) - \/12 / 2 * 2
= (6 - 2 \/3 ) / 4
= 2 (3 - \/3 ) / 4
= 3 - \/3 / 2
answer ; 3 + \/3 / 2 and 3 - \/3 / 2
(note; ^ means power)