Question

Solve the following equation: $5\sqrt {\frac{3}{x}} + 7 + \sqrt {\frac{x}{3}} = 22\frac{2}{3}.$

Answer 1

Hint: we need to know the basic factorization of quadratic equations to solve this problem.
Given equation is $5\sqrt {\frac{3}{x}} + 7 + \sqrt {\frac{x}{3}} = 22\frac{2}{3}$
For simplification of calculations put $t = \sqrt {\frac{3}{x}} $ then $\sqrt {\frac{x}{3}} = \frac{1}{t}$ , then the given equation will be
$5t + \frac{7}{t} = \frac{{68}}{3}$
Simplifying the above equation
$\frac{{5{t^2} + 7}}{t} = \frac{{68}}{3}$
$15{t^2} + 21 = 68t$
$15{t^2} - 68t + 21 = 0$
Now we got a quadratic equation, on factorization we get
$15{t^2} - 5t - 63t + 21 = 0$
$5t(3t - 1) - 21(3t - 1) = 0$
$(5t - 21)(3t - 1) = 0$
$t = \frac{{21}}{5},\frac{1}{3}$
Now, we know that $t = \sqrt {\frac{3}{x}} $
${t^2} = \frac{3}{x}$
$x = \frac{3}{{{t^2}}}$ , solving for the values of x using t value.
For $t = \frac{{21}}{5}$
$$x = \frac{3}{{{{\left( {\frac{{21}}{5}} \right)}^2}}} = \frac{{25}}{{147}}$$
For $t = \frac{1}{3}$
$x = \frac{3}{{{{\left( {\frac{1}{3}} \right)}^2}}} = 27$
$\therefore x = 27,\frac{{25}}{{147}}$ are the required values.

Note: Here we are converting the given equation into a quadratic equation by using the substitution method. We substituted $\sqrt {\frac{3}{x}} $as t, after substitution we simplified the equation and solved for t. After getting t values, we have to again substitute the value of t in terms of x, then finding x values easily.