What is x if $6 = \frac{7}{x} + x$ $6=7/x+x$ ?

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Answer 1 · 2018-05-03T04:01:53

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$6 = \frac{7}{x} + x$ $6=7/x+x$ where $x \neq 0$ $x!=0$

$\frac{7}{x} = 6 - x$ $7/x=6-x$

$x^{2} \cdot \frac{7}{x} = x^{2} (6 - x)$ $x^2*7/x=x^2(6-x)$

$7 x = 6 x^{2} - x^{3}$ $7x=6x^2-x^3$

$x^{3} - 6 x^{2} + 7 x = 0$ $x^3-6x^2+7x=0$

$x (x^{2} - 6 x + 7) = 0$ $x(x^2-6x+7)=0$

$x = 0$ $x=0$ or $x^{2} - 6 x + 7 = 0$ $x^2-6x+7=0$

For $x^{2} - 6 x + 7 = 0$ $x^2-6x+7=0$ , we need to use the quadratic formula

ie $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$

$x = \frac{6 \pm \sqrt{36 - 28}}{2}$ $x=(6+-sqrt(36-28))/(2)$

$x = \frac{6 \pm 2 \sqrt{2}}{2}$ $x=(6+-2sqrt2)/2$

$x = 3 \pm \sqrt{2}$ $x=3+-sqrt2$

BUT looking at $x = 0$ $x=0$ , it cannot be a solution because of $\frac{7}{0}$ $7/0$

Therefore, the answer is $x = 3 \pm \sqrt{2}$ $x=3+-sqrt2$

What is x if 6=7x+x6=7/x+x?