Question

How do you solve \[{\left( {x + 5} \right)^{\dfrac{1}{2}}} - {\left( {5 - 2x}
\right)^{\dfrac{1}{4}}} = 0\] and find any extraneous solutions \[?\]

Answer 1

Hint:This question involves the operation of addition/ subtraction/ multiplication/ division. Also, we need to know how to convert square root functions into power functions. We need to know how to find extraneous roots from the given equation. We need to know the basic form of a quadratic equation and the formula to find \[x\] from the quadratic equation.

Complete step by step solution:
The given equation is shown below,
\[{\left( {x + 5} \right)^{\dfrac{1}{2}}} - {\left( {5 - 2x} \right)^{\dfrac{1}{4}}} = 0 \to \left( A
\right)\]
The above equation can also be written as,
\[{\left( {x + 5} \right)^{\dfrac{1}{2}}} = {\left( {5 - 2x} \right)^{\dfrac{1}{4}}}\]
For solving the above equation we take squares on both sides of the equation.
So we get
\[{\left( {{{\left( {x + 5} \right)}^{\dfrac{1}{2}}}} \right)^2} = {\left( {{{\left( {5 - 2x}
\right)}^{\dfrac{1}{4}}}} \right)^2} \to \left( 1 \right)\]
We know that,
\[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\]
So, the equation\[\left( 1 \right)\]becomes,
\[{\left( {x + 5} \right)^{\dfrac{1}{2} \times 2}} = {\left( {5 - 2x} \right)^{\dfrac{1}{4} \times 2}}\]
So, we get
\[{\left( {x + 5} \right)^1} = {\left( {5 - 2x} \right)^{\dfrac{1}{2}}} \to \left( 2 \right)\]
Again take square on both sides of the equation\[\left( 2 \right)\], we get
\[
\left( 2 \right) \to {\left( {x + 5} \right)^1} = {\left( {5 - 2x} \right)^{\dfrac{1}{2}}} \\
{\left( {x + 5} \right)^2} = {\left( {5 - 2x} \right)^{\dfrac{1}{2} \times 2}} \\
\]
\[{\left( {x + 5} \right)^2} = \left( {5 - 2x} \right) \to \left( 3 \right)\]
We know that,
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\]
So, the equation\[\left( 3 \right)\]becomes
\[
\left( 3 \right) \to {\left( {x + 5} \right)^2} = \left( {5 - 2x} \right) \\
\left( {{x^2} + 10x + 25} \right) = \left( {5 - 2x} \right) \\
\]
The above equation can also be written as,
\[{x^2} + 25 + 10x - 5 + 2x = 0\]
By solving the above equation we get,
\[{x^2} + 12x + 20 = 0 \to \left( 4 \right)\]
The basic form of a quadratic equation is, \[a{x^2} + bx + c = 0\]
Then,
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \to \left( 5 \right)\]
By using the equation\[\left( 5 \right)\], we get the \[x\] value as follows,
\[\left( 4 \right) \to {x^2} + 12x + 20 = 0\]
Here, we have\[a = 1, b = 12\], and \[c = 20\]
So, the equation\[\left( 5 \right)\]becomes,
\[\left( 5 \right) \to x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
\[x = \dfrac{{\left( { - 12} \right) \pm \sqrt {{{\left( {12} \right)}^2} - 4 \times 1 \times 20} }}{{2
\times 1}}\]
\[
x = \dfrac{{ - 2 \pm \sqrt {144 - 80} }}{2} \\
x = \dfrac{{ - 2 \pm \sqrt {64} }}{2} \\
\]
We know that\[{8^2} = 64\]. So, the above equation\[\sqrt {64} \]can write as\[8\]. So, we get
\[

x = \dfrac{{ - 12 \pm 8}}{2} \\
x = \dfrac{{4\left( { - 3 \pm 2} \right)}}{2} \\
x = 2\left( { - 3 \pm 2} \right) \\
x = - 6 \pm 4 \\
\]
The above equation can be solved by two cases.
Case: \[1\]
We get,
\[
x = - 6 + 4 \\
x = - 2 \\
\]
Case: \[2\]
We get,
\[
x = - 6 - 4 \\
x = - 10 \\
\]
We have \[x = - 2\]and\[x = - 10\]. Let’s find the extraneous root from the values of \[x\]. Let’s
substitute \[x = - 2\]in the LHS of the equation\[\left( A \right)\], we get
LHS\[ \Rightarrow {\left( {x + 5} \right)^{\dfrac{1}{2}}}\]
\[ \Rightarrow {\left( { - 2 + 5} \right)^{\dfrac{1}{2}}} = {\left( 3 \right)^{\dfrac{1}{2}}} = \sqrt 3 \]
Let’s substitute\[x = - 2\] in the RHS of the equation \[\left( A \right)\], we get
RHS\[ \Rightarrow {\left( {5 - 2x} \right)^{\dfrac{1}{4}}}\]
\[ \Rightarrow {\left( {5 - (2 \times - 2)} \right)^{\dfrac{1}{4}}} = {\left( {5 + 4}
\right)^{\dfrac{1}{4}}} = {9^{\dfrac{1}{4}}} = {\left( 3 \right)^{2 \times \dfrac{1}{4}}} =
{3^{\dfrac{1}{2}}} = \sqrt 3 \]
Here, we get \[LHS = RHS\]
So, \[x = - 2\] is not an extraneous root of the given equation.
Let’s substitute\[x = - 10\]in the LHS of the equation\[\left( A \right)\], we get
LHS\[ \Rightarrow {\left( {x + 5} \right)^{\dfrac{1}{2}}}\]
\[ \Rightarrow {\left( { - 10 + 5} \right)^{\dfrac{1}{2}}} = {\left( { - 5} \right)^{\dfrac{1}{2}}} = \sqrt {5i} \]
Let’s substitute\[x = - 10\]in the RHS of the equation\[\left( A \right)\], we get
RHS\[ \Rightarrow {\left( {5 - 2x} \right)^{\dfrac{1}{4}}}\]
\[ \Rightarrow {\left( {5 - (2 \times - 10)} \right)^{\dfrac{1}{4}}} = {\left( {5 + 20}
\right)^{\dfrac{1}{4}}} = 9 = {25^{\dfrac{1}{4}}} = {\left( 5 \right)^{2 \times \dfrac{1}{4}}} =
{5^{\dfrac{1}{2}}} = \sqrt 5 \]
So, we get\[LHS \ne RHS\].
So, \[x = - 10\] is an extraneous root of the given equation.
So, the final answer is,
\[x = - 2\]and\[x = - 10\]
Here \[x = - 10\] is an extraneous root of the given question.

Note: Note that extraneous root means when we substitute the \[x\] values in the given equation if LHS is equal to RHS, then \[x\] the value is not an extraneous root. Otherwise, the\[x\]value is an extraneous root. Remember the formula for finding the \[x\] values from the quadratic equation to solve these types of problems. Also, this type of question involves the operation of addition/ subtraction/ multiplication/ division.