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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 \dots

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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