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# Plz help

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Find $z$ for which $\dfrac{2\sqrt{z}-3}{\sqrt{z}-1} - 1 = \dfrac{3\sqrt{z}-1}{1-\sqrt{z}}$.

Aug 10, 2022

#1
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Hi Fytonic!

Let's simplify the left-hand side first:
$$\frac{2\sqrt{z}-3-\sqrt{z}+1}{\sqrt{z}-1}=\frac{\sqrt{z}-2}{\sqrt{z}-1}$$

Now, let's equate this to the right-hand side as given:

$$\frac{\sqrt{z}-2}{\sqrt{z}-1}=\frac{3\sqrt{z}-1}{1-\sqrt{z}}$$

Now, notice: $$\frac{\sqrt{z}-2}{\sqrt{z}-1}=\frac{3\sqrt{z}-1}{-(-1+\sqrt{z})} =-\frac{3\sqrt{z}-1}{\sqrt{z}-1}$$   (we factored a negative from the denominator). So when we multiply both sides by: $$\sqrt{z}-1$$, there is no longer a denominator:

$$\implies \sqrt{z}-2=-(3\sqrt{z}-1)$$

$$\implies \sqrt{z}-2=-3\sqrt{z}+3 \\ \implies 4\sqrt{z}=5 \\ \implies \sqrt{z}=\frac{5}{4} \\ \implies z=(\frac{5}{4})^2=\frac{25}{16}$$

I hope this helps!

Aug 10, 2022
#2
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I'm sorry but it says that's wrong. Still, thanks for trying!

Fytonic  Aug 10, 2022
#3
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Oh! I am sorry, I distributed the bracket wrong in the first answer (at -(3sqrt(z) -1) I make this: -3sqrt(z)+3 while it should have been -3sqrt(z)+1.

Ok here is the correct way:

$$\dfrac{2\sqrt{z}-3-\sqrt{z}+1}{\sqrt{z}-1}=-\dfrac{3\sqrt{z}-1}{\sqrt{z}-1}$$

So:

$$\frac{\sqrt{z}-2}{1}=-3\sqrt{z}+1 \implies \sqrt{z}=-3\sqrt{z}+3 \implies 4\sqrt{z}=3 \\ \sqrt{z}=\frac{3}{4} \\ z=\frac{9}{16}$$

Aug 10, 2022
#4
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Thank you so much!

Fytonic  Aug 10, 2022