Question

Solve the equation:$\sqrt {5x - 9} - 1 = \sqrt {3x - 6}$ and find any extraneous solutions.

Answer 1

To solve this equation, we need to do squaring on both sides and after solving we need to square both sides of the equation again and on further solving we will get a quadratic equation and to find the values for $x$, we need to solve the formed quadratic equation.

Complete step by step solution:
To solve the equation, firstly we need to know that the number formed inside the square root is non negative. Let us take the value of $x$ as$1$ , then the value inside the square root becomes negative and when we take the value of $x$ equals to $2$ then the value inside the square becomes non negative and hence, we can say that the value of x is greater than or equal to 2. Now, we will solve the equation,$\sqrt {5x - 9} - 1 = \sqrt {3x - 6}$, firstly we will rewrite the equation as, $\sqrt {5x - 9} = \sqrt {3x - 6} + 1$,now, squaring both sides of the equation,
${(\sqrt {5x - 9} )^2} = {(\sqrt {3x - 6} + 1)^2}$
On squaring, we get,
$5x - 9 = 3x - 6 + 1 + 2\sqrt {3x - 6}$
On further solving, collecting like terms, we get,
$\Rightarrow 5x - 3x - 9 + 6 - 1 = 2\sqrt {3x - 6} \\\ \Rightarrow 2x - 4 = 2\sqrt {3x - 6} \\\$
Now, take $2$ common from both the sides,
$\Rightarrow 2(x - 2) = 2(\sqrt {3x - 6} )$
We get,
$\Rightarrow x - 2 = \sqrt {3x - 6}$
And, we know that $x \geqslant 2$ , so, we will square again both sides,
$\Rightarrow {(x - 2)^2} = {(\sqrt {3x - 6} )^2}$
After squaring, we get,
$\Rightarrow {x^2} + 4 - 4x = 3x - 6$
On further solving, we get,
$\Rightarrow {x^2} + 4 + 6 - 4x - 3x = 0 \\\ \Rightarrow {x^2} - 7x + 10 = 0 \\\$
Now, the equation we have, is a quadratic equation, now, we will find the values of x,
$\Rightarrow {x^2} - 7x + 10 = 0 \\\ \Rightarrow {x^2} - 2x - 5x + 10 = 0 \\\ \Rightarrow x\left( {x - 2} \right) - 5\left( {x - 2} \right) = 0 \\\ \Rightarrow (x - 5)\left( {x - 2} \right) = 0 \\\$
Now, $x - 5 = 0$ and $x - 2 = 0$ which means $x = 5$ or $x = 2$ and these are the valid solutions and the extraneous solution is $x \geqslant 2$.

Note: In this problem, we have to find the extraneous solutions and these are those solutions of an equation that comes out in the process of solving an equation but the solution that comes out is not valid. To solve the equation we had used the formula${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ and ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$.