If g(x) = sqrt((x + 3)/4), for what value of x will g(2x) = 2g(x)? Express your answer in simplest form.
+9674 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).
+9674 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).
