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ICSE Class X Math 2015 Paper

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Question : 15 of 52
Marks: +1, -0
Find the value of ' KK ' for which x=3x=3 is a solution of the quadratic equation,
(K+2)x2−Kx+6=0.(K+2) x^{2}-K x+6=0 .
Thus find the other root of the equation.
Solution:  
Given : (k+2)x2−kx+6=0(k+2) x^{2}-k x+6=0
Putting x=3x=3
(k+2)×9−k×3+6  =0(k+2) \times 9 - k \times 3 + 6\;=0
9k+18−3k+6  =09 k + 18 - 3 k + 6\;=0
6k  =−246 k\;=-24
k  =−4k\;=-4
Putting k=−4k=-4 in given equation
(−4+2)x2−(−4)x+6  =0(-4+2) x^{2} - (-4) x + 6\;=0
−2x2+4x+6  =0-2 x^{2} + 4 x + 6\;=0
x2−2x−3  =0x^{2} - 2 x - 3\;=0
x2−3x+x−3  =0x^{2} - 3 x + x - 3\;=0
x(x−3)+1(x−3)  =0x(x-3) + 1(x-3)\;=0
(x+1)(x−3)  =0(x+1)(x-3)\;=0
x+1=0     or   x−3=0x+1=0\;\;\text{ or }\; x-3=0
x=−1     or   x=3x=-1\;\;\text{ or }\; x=3
Other root of the equation, x=−1x=-1
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