EXERCISE 4.1 QUESTION 1 PART (VI),(VII),(VIII) CLASS 10T H MATHS NCERT SOLUTIONS CHPTER 4 QUADRETIC EQUATION
प्रश्न -(1) जांच कीजिये की क्या निम्न द्विघात समीकरण हे ?
(VI ) X^2 + 3X + 1 = (X -2 )^ 2
हल : X^2 + 3X + 1 = (X -2 )^ 2
X ^ 2 + 3X +1 = X ^ 2 + 2^ 2 - 2 X (2 )
X^ 2 -X^ 2 + 3X+ 4X + 1 - 4 = 0
7X -3 =0
उपरोक्त समीकरण द्विघात समीकरण नहीं हे क्युकी यहाँ पर X की अधिकतम घात 1 हे। तथा
a = 0 हे।
(vii ) (x +2 )^ 3 = 2x (x^2 -1 )
हल : (x +2 )^ 3 = 2x (x^2 -1 )
x^3 + 2^3 +3 *x^2 * 2 +3 *x* 2^ 2 =2x^3 - 2x
x^3 + 8 + 6x^2 +12x =2x^3 - 2x
x^3 -2x^3 + 6x^2 +12x +2x + 8 = 0
-x^ 3 + 6x^2 +14 x + 8 =0
(-) चिन्ह से गुणा करने पर ,
+x^ 3 - 6x^2 - 14 x - 8 =0
यह एक त्रिघात समीकरण हे।
(viii ) x^3 - 4x^2 -x +1 = (x - 2 )^ 3
हल : x^3 - 4x^2 -x +1 = (x - 2 )^ 3
x^3 - 4x^2 -x +1 = x^3 -8 -6x^ 2 +12x
2x^2 -13x +9 =0
यह एक द्विघात समीकरण हे , जो की a x^2 + b x + c = 0 प्रकार का हे। जहाँ a , शून्य नहीं हो सकता।
Question - (1) Check whether the following is a quadratic equation?
(VI ) X^2 + 3X + 1 = (X -2 )^ 2
Solution : X^2 + 3X + 1 = (X -2 )^ 2
X ^ 2 + 3X +1 = X ^ 2 + 2^ 2 - 2 X ( 2 )
X^ 2 -X^ 2 + 3X+ 4X + 1 - 4 = 0
7X -3 =0
The above equation is not a quadratic equation because here the maximum degree of X is 1.
And a = 0.
(vii ) (x +2 )^ 3 = 2x (x^2 -1 )
Solution : (x +2 )^ 3 = 2x (x^2 -1 )
x^3 + 2^3 +3 *x^2 * 2 +3 *x* 2^ 2 =2x^3 - 2x
x^3 + 8 + 6x^2 +12x =2x^3 - 2x
x^3 -2x^3 + 6x^2 +12x +2x + 8 = 0
-x^ 3 + 6x^2 +14 x + 8 =0
On multiplying by the (-) sign,
+x^ 3 - 6x^2 - 14 x - 8 =0
This is a quadratic equation.
(viii ) x^3 - 4x^2 -x +1 = (x - 2 )^ 3
Solution : x^3 - 4x^2 -x +1 = (x - 2 )^ 3
x^3 - 4x^2 -x +1 = x^3 -8 -6x^ 2 +12x
2x^2 -13x +9 =0
This is a quadratic equation of the type a x^2 + b x + c = 0 . where a cannot be zero.