Question
If a,b are real numbers such that
\sqrt{\dfrac{a}{b} } +b=7
\sqrt{\dfrac{b}{a} } +a=11
find the values of a,b
Answer 1
\sqrt{\dfrac{a}{b} } +b=7
then
\sqrt{\dfrac{a}{b} }-7=-b
(1)
\sqrt{\dfrac{b}{a} } +a=11
\sqrt{\dfrac{b}{a} } -11=-a
(2)
Dividing (1) by (2) gives
\dfrac{\sqrt{\dfrac{a}{b} }-7}{\sqrt{\dfrac{b}{a} } -11} = \dfrac{b}{a}
(3)
Let
x =\sqrt{\dfrac{a}{b} }
Then equation (3) is transformed to
\dfrac{x-7}{\dfrac{1}{x}-11 } =\dfrac{1}{x^2}
Cross-multiply
x^3-7x^2 = \dfrac{1}{x}-11
then
x^4-7x^3+11x-1 = 0
Solving the polynomial equation of degree 4 gives 4 results
\begin{cases} x_1=−1.2011606910436 (\text{cancel}) \\ x_2=6.7627135038525 \\ x_3=0.091388462397205 \\ x_4=1.347058724794 \end{cases}
Substitution results in,
When x = 6.7627135038525
\begin{cases} a = 10.85213036 \\ b = 0.237286496 \end{cases}
When x = 0.091388462
\begin{cases} a = 0.057699695 \\ b = 6.908611538 \end{cases}
When x = 1.347058724794
\begin{cases} a = 10.25764187 \\ b = 5.652941275 \end{cases}
Answer 2
From \sqrt{\dfrac{b}{a} } +a=11.
\sqrt{\dfrac{b}{a} } =11-a
Square the both sides
\dfrac{b}{a} = (11-a)^2
Substituting to \sqrt{\dfrac{a}{b} } +b=7 eliminate the b from the equation, which is ended up with a quartic equation.
a^4-33a^3+363a^2-1332a+76 = 0
Solving the quartic equation results in the same outcome as that obtained by the method 1