Chapter 1.5, Problem 4E

(a)

To determine

To isolate: The radical in the equation $\sqrt{2x}+x=0$.

Answer to Problem 4E

The radical in the equation $\sqrt{2x}+x=0$ can be isolated as $\underset{\_}{\sqrt{2x}=-x}$.

Explanation of Solution

Consider the given equation $\sqrt{2x}+x=0$.

The term involving the radical sign is on the left-hand side of the equation.

Therefore, subtract $x$ from both sides of the above equation to isolate the radical.

$\begin{array}{l}\sqrt{2x}+x-x=0-x\\ \sqrt{2x}=-x\end{array}$

Thus, the radical in the equation $\sqrt{2x}+x=0$ can be isolated as $\underset{\_}{\sqrt{2x}=-x}$.

(b)

To determine

To square: Both sides of the equation $\sqrt{2x}=-x$.

Answer to Problem 4E

Both sides of the equation $\sqrt{2x}+x=0$ can be squared as $\underset{\_}{2x=x^2}$.

Explanation of Solution

Consider the equation $\sqrt{2x}=-x$ from part (a).

To solve this equation, the radical sign has to be eliminated.

Therefore, square both sides of the above equation.

$\begin{array}{l}{\left(\sqrt{2x}\right)}^2={\left(-x\right)}^2\\ 2x=x^2\end{array}$

Thus, both sides of the equation $\sqrt{2x}+x=0$ can be squared as $\underset{\_}{2x=x^2}$.

(c)

To determine

The solutions of the equation $2x=x^2$.

Answer to Problem 4E

The solutions of the equation $2x=x^2$ are $\underset{\_}{x=0\text{or }x=2}$.

Explanation of Solution

Formula used:

Quadratic formula:

The solution of a quadratic equation of the form $a{x}^2+bx+c=0,a\ne 0$ can be obtained by using the quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

Calculation:

The equation $2x=x^2$ from part (b) can be rewritten as $x^2-2x=0$.

Compare this equation with the general form $a{x}^2+bx+c=0$.

Here $a=1,b=-2\text{and }c=0$.

Substitute 1 for a, $-2$ for b and $0$ for c in the quadratic formula to find the solution.

$\begin{array}{c}x=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(0\right)}}{2\cdot 1}\\ =\frac{2\pm \sqrt{4}}{2}\\ =\frac{2\pm 2}{2}\\ =1+1\text{or }1-1\\ =2\text{or 0}\end{array}$

Thus, the solutions of the equation $2x=x^2$ are $\underset{\_}{x=0\text{or }x=2}$.

(d)

To determine

The solutions of the equation $2x=x^2$ that satisfy the original equation $\sqrt{2x}+x=0$.

Answer to Problem 4E

The solution of the equation $2x=x^2$ that satisfy the original equation $\sqrt{2x}+x=0$ is $\underset{\_}{x=0}$.

Explanation of Solution

From part (c) the solutions of the equation $2x=x^2$ are $x=0\text{or }x=2$.

To check whether these solutions satisfy the original equation, substitute these values in the equation $\sqrt{2x}+x=0$.

Substitute 0 for x in $\sqrt{2x}+x=0$.

$\begin{array}{c}\sqrt{2x}+x=0\\ \sqrt{2\cdot 0}+0=0\\ 0=0\left[\text{LHS}=\text{RHS}\right]\end{array}$

Substitute 2 for x in $\sqrt{2x}+x=0$.

$\begin{array}{c}\sqrt{2x}+x=0\\ \sqrt{2\cdot 2}+2=0\\ \sqrt{4}+2=0\\ 4=0\left[\text{LHS}\ne \text{RHS}\right]\end{array}$

The value $x=2$ does not satisfy the original equation. Therefore, it is not a solution of the equation $\sqrt{2x}+x=0$.

Thus, the solution of the equation $2x=x^2$ that satisfy the original equation $\sqrt{2x}+x=0$ is $\underset{\_}{x=0}$.