Calculate $f(x)$ at a specific point

Question

Question: Calculate $f(x) = \frac{49}{x^2} + x^2$ at points for which $\frac{7}{x}+x =3$

My attempt:- I tried to find the value of $x$ and insert in $f(x)$ $$\frac{7}{x}+x =3$$ $$7+x^2 =3x$$ $$x^2 -3x + 7=0$$

$$x = \frac{3+\sqrt{9-7*4}}{2}$$

Now $x$ is coming out to be irrational and things get a lot more difficult from there. What should I do?

Answer 1

Hint:

You can use the following identity:-

$$(\frac{1}{a}-a)^2 = \frac{1}{a^2} + a^2 -2 $$

Answer 2

you Can plug in $$x=\frac{1}{2}(3 \pm i\sqrt{19})$$ into your formula for your Control: the result is $-5$

Answer 3

For $x \not =0$, $x$ real:

1) No real, negative $x$ satisfies above equation.

2) For $x >0$ :AM-GM:

$(7/x) +x \ge 2\sqrt{7} \gt 4 $.

Hence there is no real $x$ with: $(7/x) +x = 3$ .

Left to do, find a complex solution.

Note:

$(7/x+x)^2 =$

$ 49/x^2 +2(7) + x^2 =9.$

$49/x^2 +x^2 =-5.$

Hence $f(x)=-5.$

Answer 4

Make use of $$ \left(\frac{7}{x}+x \right)^2=\frac{49}{x^2}+x^2+14 $$ i.e. $$ 3^2-14=\frac{49}{x^2}+x^2=-5 $$