Question

How do You solve \[{{\left( 5x+4 \right)}^{{}^{1}/{}_{2}}}-3x=0,\]and find any extraneous solution.

Answer 1

Hint: In the given question, we have been asked to check an extraneous solution. For this we must know what an extraneous solution is. An extraneous solution to an equation is a solution which is obtained by solving with appropriate method, but the solution is not fit for the equation.

Complete step by step solution:
The given equation is
\[{{\left( 5x+4 \right)}^{{}^{1}/{}_{2}}}-3x=0\]
Which can be written as
\[\sqrt{5x+4}-3x=0\]
\[\therefore \sqrt{5x+4}=3x\]
Now take the square on both sides.
\[\Rightarrow 5x+4=9{{x}^{2}}\]
\[\therefore 9{{x}^{2}}-5x-4=0\]
\[\Rightarrow 9{{x}^{2}}-9x+4x-4=0\]
\[\Rightarrow (9x+4)(x-1)=0\]
Hence, the solution for \[x\]are
\[\Rightarrow \]\[x=1,\dfrac{-4}{9}\]
Now, check these values in original equation.
\[(i)\,\,x=1\]
\[\Rightarrow (5x+4){}^{1}/{}_{2}-3x=0\]
\[\Rightarrow \left( 5+4 \right){}^{1}/{}_{2}-3=0\]
\[\Rightarrow 0=0\]
\[\therefore \]It’s a true solution.
\[(ii)\,\,x=-\dfrac{4}{9}\]
\[\Rightarrow {{\left( 5x+4 \right)}^{{}^{1}/{}_{2}}}-3x=0\]
\[\Rightarrow {{\left( 5\left( -\dfrac{4}{9} \right)+4 \right)}^{{}^{1}/{}_{2}}}-3\left( -\dfrac{4}{9} \right)=0\]
\[\Rightarrow {{\left( -\dfrac{20}{9}+\dfrac{36}{9} \right)}^{{}^{1}/{}_{2}}}+\dfrac{12}{9}=0\]
\[\Rightarrow \dfrac{4}{3}+\dfrac{4}{3}=0\]
\[\Rightarrow \dfrac{8}{3}=0\]
\[\therefore \,\,\] It’s not a true solution.

Hence, \[x=-\dfrac{4}{9}\] is an extraneous solution.

Note: In the given question, we had to check for extraneous solutions. Always remember to check for this is every quadratic equation or equations with more than one solutions, that if they break any fundamental rule for example, in this question we don’t get zero as an answer for \[x=-\dfrac{4}{9}.\]