Show that the function $g (x) = \sqrt{x - 5} - \frac{1}{x + 3}$ $g(x)=sqrt(x-5)-1/(x+3)$ has at least a real root?

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Answer 1 · 2018-04-11T13:42:11

$x = 5$ $color(blue)(x=5)$

$\sqrt{x - 5} - \frac{1}{x + 3} = 0$ $sqrt(x-5)-1/(x+3)=0$

Add $\frac{1}{x + 3}$ $1/(x+3)$

$\sqrt{x - 5} = \frac{1}{x + 3}$ $sqrt(x-5)=1/(x+3)$

Squaring:

$x - 5 = \frac{1}{{(x + 3)}^{2}}$ $x-5=1/(x+3)^2$

Multiply by ${(x + 3)}^{2}$ $(x+3)^2$

${(x + 3)}^{2} (x - 5) = 1$ $(x+3)^2(x-5)=1$

${(x + 3)}^{2} (x - 5) - 1 = 0$ $(x+3)^2(x-5)-1=0$

Factor:

$(x - 5) [{(x + 3)}^{2} - \frac{1}{x - 5}] = 0$ $(x-5)[(x+3)^2-1/(x-5)]=0$

$x - 5 = 0 \Rightarrow x = 5$ $x-5=0=>color(blue)(x=5)$

Show that the function g(x)=√x−5−1x+3g(x)=sqrt(x-5)-1/(x+3) has at least a real root?