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Question Number 136344 by Gideon last updated on 21/Mar/21

Commented by mindispower last updated on 21/Mar/21

$$x^2+6x={(x+3)}^2-9$$$$t={(x+3)}^2$$$$\frac{5}{\sqrt{t-1}}+\frac{1}{\sqrt{t-4}}=\frac{4}{t}$$$$wehavet>4,t-1>3\Rightarrow \sqrt{t-1}<t\iff$$$$\frac{5}{\sqrt{t-1}}>\frac{4}{\sqrt{t-1}}>\frac{4}{t}$$$$nosolutionover\mathbb{R}$$$$$$

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