Explains Fraction Equations Reducible to Quadratic for Free Math Help.
\( \begin{aligned} \displaystyle
x+\frac{a}{x} &= b \\
x^2 + a &= bx \\
x^2-bx + a &= 0
\end{aligned} \)
Once the equation forms a quadratic form by multiplying the denominator by both sides, then the equation can be solved by a quadratic solution, such as factories or the quadratic formula.
Let’s look at the following examples.
Practice Questions
Question 1
Solve \( \displaystyle x+\frac{2}{x} = 3 \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
x^2 + 2 &= 3x &\color{red} \text{multiply both sides by } x \\
x^2-3x + 2 &= 0 \\
(x-1)(x-2) &= 0 \\
\therefore x = 1 &\text{ or } x=2
\end{aligned} \)
Question 2
Solve \( \displaystyle x^3-\frac{8}{x^3} = 7 \).
\( \begin{aligned} \require{AMSsymbols} \displaystyle
x^6-8 &= 7x^3 &\color{red} \text{multiply both sides by } x^3 \\
x^6-7x^3-8 &= 0 \\
(x^3+1)(x^3-8) &= 0 \\
x^3 = -1 &\text{ or } x^3 = 8 \\
\therefore x = -1 &\text{ or } x = 2
\end{aligned} \)
Question 3
Solve \( \displaystyle 16x^2 + \frac{16}{x^2} = 257 \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
16x^4 + 16 &= 257x^2 &\color{red} \text{multiply both sides by } x^2 \\
16x^4-257x^2 + 16 &= 0 \\
(16x^2-1)(x^2-16) &= 0 \\
(4x-1)(4x+1)(x-4)(x+4) &= 0 \\
\therefore x &= \frac{1}{4}, -\frac{1}{4}, 4, -4
\end{aligned} \)
Question 4
Solve \( \displaystyle \sqrt{x} + \frac{1}{\sqrt{x}} = 2 \).
\( \begin{aligned} \displaystyle \require{AMSsymbols} \require{color}
\bigg(\sqrt{x} + \frac{1}{\sqrt{x}}\bigg)^2 &= 2^2 &\color{red} \text{square both sides} \\
x + 2 + \frac{1}{x} &= 4 \\
x-2 + \frac{1}{x} &= 0 \\
x^2-2x + 1 &= 0 &\color{red} \text{multiply both sides by } x \\
(x-1)^2 &= 0 \\
\therefore x&= 1
\end{aligned} \)
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