Grade 12Algebra

Sir.. m nt getting this que.. can u help me ..... i tried but ans. Getting 2 bt real ans is root 2 and root 6

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Mithu

10 Years agoGrade 12

1 Answer

Ajay

10 Years ago

Neither of the answer seems to be correct. 0 and 7/4 are possible values for x as equations is satosfied for these values. There are two more more solutions which I have explained below.

$\large Put\quad y\quad =\quad log(4-x)\quad and\quad a\quad =\quad log(x+1/2)\quad the\quad equation\quad reduces\quad to\\ { y }^{ 2 }+ay-2{ a }^{ 2 }\quad =\quad 0\quad or\quad (y-a)(y+2a)\quad =\quad 0.\\ y=\quad a\quad or\quad y\quad =-2a\\ log(4-x)\quad =\quad log(x+1/2)-----(1)\\ log(4-x)\quad =\quad -2log(x+1/2)-----(2)\\ Solving\quad (1)\quad we\quad get\quad x=7/4\quad as\quad one\quad solution.\\ Solving\quad (2)\quad we\quad get\\ log(4-1){ (x+1/2) }^{ 2 }\quad =\quad 0\\ or\quad (4-1){ (x+1/2) }^{ 2 }\quad =\quad 1,\quad \\ This\quad reduces\quad to\quad x(4{ x }^{ 2 }-12x-15)\quad =\quad 0\\ hence\quad x\quad =\quad 0\quad is\quad second\quad solution\\ Solving\quad 4{ x }^{ 2 }-12x-15=\quad 0\quad will\quad give\quad two\quad more\quad solution$