Let f(x)=x^2 −2x−3; x≥1 & g(x)=1+(√(x+4)) ; x≥−4 then the number of real solutions of equation f(x)=g(x) is ...


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Question Number 137817 by bramlexs22 last updated on 07/Apr/21

$L e t f (x) = x^{2} - 2 x - 3; x ⩾ 1 &$ $g (x) = 1 + \sqrt{x + 4}; x ⩾ - 4 t h e n$ $t h e n u m b e r o f r e a l s o l u t i o n s$ $o f e q u a t i o n f (x) = g (x) i s . . .$
	Answered by MJS_new last updated on 07/Apr/21

	$x^{2} - 2 x - 3 = 1 + \sqrt{x + 4}$ $x^{2} - 2 x - 4 = \sqrt{x + 4}$ ${(x^{2} - 2 x - 4)}^{2} = x + 4$ $x^{4} - 4 x^{3} - 4 x^{2} + 15 x + 12 = 0$ $I found no ‘ ‘ {nice}^{″} exact solution$ $of the 4 solutions only 2 solve the given equation$ $x_{1} \approx - 1.56155$ $x_{2} \approx 3.79129$ $but because of x ⩾ 1 the only solution is x_{2}$
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