Question

If √x/y + √y/x = 10/3 , then find xy
[HINT : square both the sides]​

Answer 1

Given to solve: $\sqrt{\dfrac{x}{y} } +\sqrt{\dfrac{y}{x} } =\dfrac{10}{3}$ for $\dfrac{x}{y}$.

We spot symmetry in two radicands. Let's substitute the radicals for a value so that we work out the restrictions later.

$t=\sqrt{\dfrac{x}{y} }$ then given equation becomes $t+\dfrac{1}{t} =\dfrac{10}{3}$.

$\Longleftrightarrow 3t^2-10t+3=0$

$\Longleftrightarrow (3t-1)(t-3)=0$

Hence $t=\dfrac{1}{3}$ or $t=3$.

Then $\sqrt{\dfrac{x}{y} } =\dfrac{1}{3}$ or $\sqrt{\dfrac{x}{y} } =3$.

The solutions are $\dfrac{x}{y} =\dfrac{1}{9}$ or $\dfrac{x}{y} =9$.

More information:

Equations with symmetry and two cases.

Symmetrical equations are one of the examples we can solve using substitution.

For example, we have $x^4-8x^3+9x^2-8x+1=0$.

The coefficients form symmetry against the middle. How do we solve this?

For even number highest coefficient.

Divide both sides by $x^2$

$\Longleftrightarow x^2-8x+9-\dfrac{8}{x} +\dfrac{1}{x^2} =0$

The substitution $t=x+\dfrac{1}{x}$ makes this a quadratic equation.

• $t=x+\dfrac{1}{x}$ then $t^2-2=x^2+\dfrac{1}{x^2}$.

Now we rearrange the terms.

$(x^2+\dfrac{1}{x^2} )-8(x+\dfrac{1}{x} )+9=0$

$\Longleftrightarrow (t^2-2)-8t+9=0$

$\Longleftrightarrow t^2-8t+7=0$

$\Longleftrightarrow (t-1)(t-7)=0$

$\Longleftrightarrow (x+\dfrac{1}{x} -1)(x+\dfrac{1}{x} -7)=0$

Solving this gives four distinct roots.

$x=\dfrac{1\pm\sqrt{3}i }{2} ,\dfrac{7\pm3\sqrt{5} }{2}$

For odd number highest coefficient.

What if we solve $x^5-7x^4+x^3+x^2-7x+1=0$ using the same logic? It forms symmetry as well as the previous one.

But it does not form symmetry against the middle.

It requires one extra step. It is factor theorem.

We observe $x=-1$ as a solution. Then it has $x+1$ as a factor.

Using synthetic division

$\Longleftrightarrow (x+1)(x^4-8x^3+9x^2-8x+1)=0$

Then we can solve this equation using the previous solution.