Question

Other Equations Find all real solutions of the equation.

$x^{\frac{1}{2}}+3x^{-\frac{1}{2}}=10x^{-\frac{3}{2}}$

Short answer

The real solution of the given equation is $x=2$.

Step-by-step solution

Step 1. Definition of equation.

Equation is a mathematical statement in which two expressions are equal to each other.

Step 2. Given Information.

The given equation is:

$x^{\frac{1}{2}}+3x^{-\frac{1}{2}}=10x^{-\frac{3}{2}}$

Step 3. Simplify the given equation.

The given equation can be rewritten as:

$\begin{array}{c}{x}^{\frac{1}{2}}+\frac{3}{x^{\frac{1}{2}}}=\frac{10}{x^{\frac{3}{2}}}\\ \frac{x+3}{x^{\frac{1}{2}}}=\frac{10}{x^{\frac{3}{2}}}\end{array}$

LCM of$x^{\frac{1}{2}}$ and$x^{\frac{3}{2}}$ is $x^{\frac{3}{2}}$, so multiply both sides by$x^{\frac{3}{2}}$ to simplify the given equation.

role="math" localid="1643912635073" $$\begin{gathered} x^{\frac{3}{2}}\times \frac{x+3}{x^{\frac{1}{2}}}=x^{\frac{3}{2}}\times \frac{10}{x^{\frac{3}{2}}} \\ x\left(x+3\right)=10 \\ {x}^2+3x-10=0 \end{gathered}$$

Step 4. Solve the above equation.

In the above equation$3$ can be written as$5-2$ because the sum of$5$ and$-2$ is$3$ and their product is $-10$.

role="math" localid="1643912791830" $\begin{array}{c}{x}^2+\left(5-2\right)x-10=0\\ x^2+5x-2x-10=0\\ x\left(x+5\right)-2\left(x+5\right)=0\\ \left(x-2\right)\left(x+5\right)=0\end{array}$

Use the zero product property.

$\begin{array}{c}x-2=0\\ x=2\end{array}$

And,

role="math" localid="1643912875418" $$\begin{gathered} x+5=0 \\ x=-5 \end{gathered}$$

The value$x^{\frac{1}{2}}$ is not defined for negative values. So,$x=-5$ cannot be a solution.

Thus, the real solution of the given equation is $x=2$.