Question

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Answer if you know!!!!

Find the value of K for which x=3 is a solution of the quadratic equation, (K+2)x²-Kx+6=0. Find the other root of the equation.
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Answer 1

$\sf \Large Solution!!$

$\sf \large Given:$

$\sf \to x=3\;is\;a\;solution\;of\;(K+2)x^{2}-Kx+6=0$

$\sf \large To\;find:$

$\sf \to Other\;root\;of\;equation.$

$\sf \large So,$

$\sf \to (K+2)\times 3^{2}-K\times 3+6=0$

$\sf \to (K+2)\times 9-K\times 3+6=0$

$\sf \to 9K+18-3K+6=0$

$\sf \to 6K+24=0$

$\sf \to 6K=-24$

$\sf \to K=-4$

$\sf \large Now,$

$\sf For\;K=-4,$

$\sf \to (-4+2)\times x^{2}-(-4)\times x+6=0$

$\sf \to -2x^{2}+4x+6=0$

$\sf \to x^{2}-2x-3=0$

$\sf \to x^{2}-3x+x-3=0$

$\sf \to x(x-3)+1(x-3)=0$

$\sf \to (x-3)(x+1)=0$

$\sf \to x=3\;or\;x=-1$

Since, x=3 is already given to be one root(solution) of the equation.

$\boxed{\therefore The\;other\;root\;of\;the\;equation\;is\;x=-1}$

Answer 2

Hope this will help you
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