#### 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

Collected in the board: Quartic Equations

Steven Zheng posted 3 months ago

\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}

Steven Zheng posted 3 months ago

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

Steven Zheng posted 8 hours ago

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