MHB What is the Solution to 3x^(-1/2) - 4 = 0?

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To solve the equation 3x^(-1/2) - 4 = 0, the first step involves multiplying through by x^(1/2) to eliminate the negative exponent. This results in the equation simplifying to 3 - 4x^(1/2) = 0. The next step is to isolate x^(1/2) by rearranging the equation to 4x^(1/2) = 3. Finally, squaring both sides leads to the solution x = (3/4)^2, which simplifies to x = 9/16. The discussion emphasizes the importance of manipulating exponents correctly to solve the equation.
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solve the equation
3x^(-1/2) - 4 = 0
 
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I've moved this thread here to our algebra forum since this is a better fit given the question.

We are given to solve:

$$3x^{-\frac{1}{2}}-4=0$$

What do we get if we multiply through by $x^{\frac{1}{2}}\ne0$?
 
MarkFL said:
I've moved this thread here to our algebra forum since this is a better fit given the question.

We are given to solve:

$$3x^{-\frac{1}{2}}-4=0$$

What do we get if we multiply through by $x^{\frac{1}{2}}\ne0$?

i don't get it?
 
Dil said:
i don't get it?

Well if we multiply through by $x^{\frac{1}{2}}$ we have:

$$3x^{-\frac{1}{2}}x^{\frac{1}{2}}-4x^{\frac{1}{2}}=0x^{\frac{1}{2}}$$

Now, for the first term on the left, we can use the following property of exponents:

$$a^{b}\cdot a^{c}=a^{b+c}$$

So, what does this term become?
 

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