NCERT SOLUTIONS FOR CLASS 10TH MATHS EXERCISE 4.3 QUESTION NO. (3) PART (I),(II)
प्रश्न -(1) निम्न समीकरणों के मूल ज्ञात कीजिये।
(I ) X - 1 / X = 3 , जहाँ X शून्य के बराबर नहीं हे।
हल : X^2 - 3X - 1 = 0
यहाँ , a = 1 , b = -3 , c = -1
अब धन चिन्ह लेने पर ,
x = {-b + \/b^2 - 4ac } / 2a
x = { -(-3) + \/(-3 )^2 - 4*1*(-1 ) } / 2 * 1
x = {3 + \/9 +4} / 2
x = { 3 + \/13 } / 2
अब ऋण चिन्ह लेने पर ,
x = {-b - \/b^2 - 4*a *c } / 2 *a
x = { - (-3 ) - \/(-3 )^2 - 4 *1 *(-1 ) } / 2 * 1
x = { 3 - \/9 + 4 } / 2
x = { 3 - \/13 } / 2
अतः अभीष्ट मूल x = { 3 + \/13 } / 2 , और x = { 3 - \/13 } / 2 :उत्तर
(ii)
Question - (1) Find the roots of the following equations.
(I) X – 1 / X = 3, where X is not equal to zero.
Solution : X^2 - 3X - 1 = 0
Here, a = 1 , b = -3 , c = -1
Now taking the plus sign,
x = {-b + \/b^2 - 4ac } / 2a
x = { -(-3) + \/(-3 )^2 - 4*1*(-1 ) } / 2 * 1
x = {3 + \/9 +4} / 2
x = { 3 + \/13 } / 2
Now taking the minus sign,
x = {-b - \/b^2 - 4*a *c } / 2 *a
x = { - (-3 ) - \/(-3 )^2 - 4 *1 *(-1 ) } / 2 * 1
x = { 3 - \/9 + 4 } / 2
x = { 3 - \/13 } / 2
So the required root x = { 3 + \/13 } / 2 , and x = { 3 - \/13 } / 2 :Answer
(II )
answer : x = 1 , 2