Question

NEED ANSWER ASAP -50 POINTS

Answer 1

$x+\dfrac{3}{x}=7$

This is a disguised quadratic equation. To solve it, we must first multiply each term by x to get rid of the fraction:

$x\times x+\dfrac{3}{x} \times x=7 \times x$

$x^2+3=7x$

Now, we could solve this equation like a normal quadratic. Move all terms to the left side of the equation:

$x^2-7x+3=0$

To solve for x, we could use the quadratic formula:

$x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}$ where the given equation is $ax^2+bx+c=0$

Given $x^2-7x+3=0$, we know that:

a = 1

b = -7

c = 3

Plug in a, b and c:

$x=\dfrac{-(-7) \pm \sqrt{(-7)^2-4(1)(3)}}{2(1)}\\\\x=\dfrac{7 \pm \sqrt{49-12}}{2}\\\\x=\dfrac{7 \pm \sqrt{37}}{2}$

Therefore, the two solutions for x are $x=\dfrac{7 - \sqrt{37}}{2}$ and $x=\dfrac{7 + \sqrt{37}}{2}$.

I hope this helps!