Answer to Question #246288 in Algebra for Adawiyah

Question

5÷(√(x+1))-3√(x+1)+2=0

Answer 1

Let $t=\sqrt{x+1}, t\geq0.$ Then

$$\dfrac{5}{t}-3t+2=0$$

$$3t^2-2t-5=0$$

$$D=(-2)^2-4(3)(-5)=64$$

$$t=\dfrac{2\pm\sqrt{64}}{2(3)}=\dfrac{1\pm4}{3}$$

Since $t\geq0,$ we take $t=\dfrac{1+4}{3}=\dfrac{5}{3}.$

Then

$$\sqrt{x+1}=\dfrac{5}{3}$$

$$x+1=\dfrac{25}{9}$$

$$x=\dfrac{16}{9}$$