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If g(x) = sqrt((x + 3)/4), for what value of x will g(2x) = 2g(x)? Express your answer in simplest form.

 Apr 16, 2022

Best Answer 

 #1
avatar+9459 
+1

We first need to find out an explicit form of g(2x).

 

\(\begin{array}{rcl} g(x) &=& \sqrt{\dfrac{x + 3}{4}} \end{array}\)

Replace x by 2x:

\(\begin{array}{rcl} g(2x) &=& \sqrt{\dfrac{2x+ 3}{4}} \end{array}\)

 

Now we know what g(2x) and g(x) are, we proceed to solve the equation.

 

\(\begin{array}{rcl} g(2x) &=& 2g(x)\\ \sqrt{\dfrac{2x + 3}4} &=& 2\sqrt{\dfrac{x + 3}{4}}\\ \dfrac{\sqrt{2x + 3}}2 &=& \sqrt{x + 3}\\ \left(\dfrac{\sqrt{2x + 3}}2\right)^2 &=& \left(\sqrt{x + 3}\right)^2\\ \dfrac{2x + 3}4&=&x + 3\\ 2x+3&=&4(x+3)\\ 2x+3&=&4x+12\\ -9&=&2x\\ x &=& -\dfrac92 \end{array}\)

 

However, when we substitute \(x =-\dfrac92\) into the equation, we see that \(x = -\dfrac92\) is not in the domain of g(x).

Hence, it is an extraneous root and needs to be rejected.

There is no such value of x that satisfies g(2x) = 2g(x).

 Apr 16, 2022
 #1
avatar+9459 
+1
Best Answer

We first need to find out an explicit form of g(2x).

 

\(\begin{array}{rcl} g(x) &=& \sqrt{\dfrac{x + 3}{4}} \end{array}\)

Replace x by 2x:

\(\begin{array}{rcl} g(2x) &=& \sqrt{\dfrac{2x+ 3}{4}} \end{array}\)

 

Now we know what g(2x) and g(x) are, we proceed to solve the equation.

 

\(\begin{array}{rcl} g(2x) &=& 2g(x)\\ \sqrt{\dfrac{2x + 3}4} &=& 2\sqrt{\dfrac{x + 3}{4}}\\ \dfrac{\sqrt{2x + 3}}2 &=& \sqrt{x + 3}\\ \left(\dfrac{\sqrt{2x + 3}}2\right)^2 &=& \left(\sqrt{x + 3}\right)^2\\ \dfrac{2x + 3}4&=&x + 3\\ 2x+3&=&4(x+3)\\ 2x+3&=&4x+12\\ -9&=&2x\\ x &=& -\dfrac92 \end{array}\)

 

However, when we substitute \(x =-\dfrac92\) into the equation, we see that \(x = -\dfrac92\) is not in the domain of g(x).

Hence, it is an extraneous root and needs to be rejected.

There is no such value of x that satisfies g(2x) = 2g(x).

MaxWong Apr 16, 2022

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