X=3-√5 find the value of √x+ 1/√x

1 answer:

5 0

$\bf x = 3-\sqrt{5} \\\\[-0.35em] ~\dotfill\\\\ \sqrt{x}+\cfrac{1}{\sqrt{x}}\implies (3-\sqrt{5})+\cfrac{1}{3-\sqrt{5}}\qquad \impliedby \textit{let use the LCD of }3-\sqrt{5} \\\\\\ \cfrac{(3-\sqrt{5})3-(3-\sqrt{5})\sqrt{5}+(1)1}{3-\sqrt{5}} \implies \cfrac{9-3\sqrt{5}-(3\sqrt{5}-5)+1}{3-\sqrt{5}} \\\\\\ \cfrac{9-3\sqrt{5}-3\sqrt{5}+5+1}{3-\sqrt{5}}\implies \cfrac{15-6\sqrt{5}}{3-\sqrt{5}}$

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Please help!!!!!>>>>>>

12345 [234]

Answer:

-3.3

Step-by-step explanation:

-2.3-(4.5-3 1/2)=-2.3-(4.5-3.5)=-2.3-(1)=-3.3

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7 0

3 years ago

(X^4)-(45x^2)+324 what are the zeroes? And how many turning points?

dmitriy555 [2]

Zeroes:

We must solve

$x^4-45x^2+324=0$

To do so, we define the auxiliary variable $t=x^2$ . The equation becomes

$t^2-45t+324=0$

The quadratic formula yields the solutions

$t=9,\quad t=36$

Substituting back $t=x^2$ gives

$x^2=9 \iff x=\pm 3,\quad x^2=36 \iff x=\pm 6$

So, the zeroes are -6, -3, 3, 6.

Turning points:

Turning points are points where a function stops being increasing to become decreasing, or vice versa. Since functions are increasing when their first derivative is positive and decreasing when it's negative, turning points are points where the first derivative is zero.

We have

$f(x)=x^4-45x^2+324 \implies f'(x)=4x^3-90x$