[tex]x+\dfrac{3}{x}=7[/tex]
This is a disguised quadratic equation. To solve it, we must first multiply each term by x to get rid of the fraction:
[tex]x\times x+\dfrac{3}{x} \times x=7 \times x[/tex]
[tex]x^2+3=7x[/tex]
Now, we could solve this equation like a normal quadratic. Move all terms to the left side of the equation:
[tex]x^2-7x+3=0[/tex]
To solve for x, we could use the quadratic formula:
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex] where the given equation is [tex]ax^2+bx+c=0[/tex]
Given [tex]x^2-7x+3=0[/tex], we know that:
a = 1
b = -7
c = 3
Plug in a, b and c:
[tex]x=\dfrac{-(-7) \pm \sqrt{(-7)^2-4(1)(3)}}{2(1)}\\\\x=\dfrac{7 \pm \sqrt{49-12}}{2}\\\\x=\dfrac{7 \pm \sqrt{37}}{2}[/tex]
Therefore, the two solutions for x are [tex]x=\dfrac{7 - \sqrt{37}}{2}[/tex] and [tex]x=\dfrac{7 + \sqrt{37}}{2}[/tex].
I hope this helps!
