Question

If x = 3 - V7, then find the value of x + 1/x.
Select one:
O a. 1/2 (7-V10)
b. 1/2 (9-17)
O c. 1/2 (7 + V10)
O d. 1/2 (9 + 17)​

Answer 1

Answer:

We can find the value of $x+\frac{1}{x}$ using the given $x$ value.

Final Answer: Option b: $x+\frac{1}{x} =\frac{1 }{2}(9- \sqrt{7})$

Step-by-step explanation:

From the given question,

The value of $x=3-\sqrt{7}$

                     $\frac{1}{x}=\frac{1}{3-\sqrt{7} }$

Step 1:

• To find the value of $\frac{1}{x}$ take the conjugate of the denominator value.
• Conjugate : Changing the sign between two terms.
• For example, the conjugate of $x + y$ is $x-y$

                    $\frac{1}{x}=\frac{1}{3-\sqrt{7} }$

                       $=\frac{1}{3-\sqrt{7} }\times \frac{3+\sqrt{7} }{3+\sqrt{7} }$

                       $=\frac{3+\sqrt{7} }{(3-\sqrt{7} )(3+\sqrt{7}) }$

Step 2:

• Denominator in the form of $(a+b)(a-b)$.
• Formula:  $(a+b)(a-b)=a^2-b^2$
• Apply this formula in step 1.So that,

                  $(3-\sqrt{7} )(3+\sqrt{7} )=3^2-(\sqrt{7} )^2$

                                              $=9-7$  => $(\sqrt{7} \times \sqrt{7} =7)$

                 $(3-\sqrt{7} )(3+\sqrt{7} )=2$

Step 3:

• Now substitute  $(3-\sqrt{7} )(3+\sqrt{7} )=2$ in step 1.

                  $\frac{1}{x} =\frac{3+\sqrt{7} }{(3-\sqrt{7} )(3+\sqrt{7}) }$

                  $\frac{1}{x}= \frac{3+\sqrt{7} }{2 }$

Step 4:

• Now we can find the value of $x+\frac{1}{x}$ .

            $x+\frac{1}{x}=(3-\sqrt{7})+\frac{3+\sqrt{7} }{2}$

• Take the L.C.M for denominator values 1 and 2
• L.C.M of 1 and 2 is 2.
• So we should equate to 2 in the denominator.That is,

                     $=\frac{2(3-\sqrt{7})+(3+\sqrt{7}) }{2}$        

                     $=\frac{(6-2 \sqrt{7})+(3+\sqrt{7}) }{2}$

                     $=\frac{6-2 \sqrt{7}+3+\sqrt{7} }{2}$

Step 5:

• We can add the terms in the numerator.

                     $=\frac{9- \sqrt{7} }{2}$

• It can be written as,

           $x+\frac{1}{x} =\frac{1 }{2}(9- \sqrt{7})$

Final Answer:

Option b:  $x+\frac{1}{x} =\frac{1 }{2}(9- \sqrt{7})$

                 

Answer 2

We need to recall the following definition of conjugate.

Conjugate is a change in the sign in the middle of two terms.

($+$ to $-$ ) or ($-$ to $+$)

This problem is about the conjugation of a term.

Given:

$x=3-\sqrt{7}$

Now,

$x+\frac{1}{x}$

$=(3-\sqrt{7} )+\frac{1}{(3-\sqrt{7} )}$

$=\frac{(3-\sqrt{7} )*(3-\sqrt{7} )+1}{(3-\sqrt{7} )}$

$=\frac{9-6\sqrt{7} +7+1}{(3-\sqrt{7} )}$

$=\frac{17-6\sqrt{7}}{3-\sqrt{7} }$

Multiply the numerator and denominator by the conjugate of $(3-\sqrt{7})$.

Conjugate of $(3-\sqrt{7})$ is $(3+\sqrt{7})$.

$=\frac{(17-6\sqrt{7})(3+\sqrt{7})}{(3-\sqrt{7})(3+\sqrt{7}) }$

$=\frac{51-18\sqrt{7}-42+17\sqrt{7}}{9-7 }$

$=\frac{9-\sqrt{7} }{2}$

$=\frac{1}{2}(9-\sqrt{7} )$

Hence, the correct option is (b) $\frac{1}{2}(9-\sqrt{7})$.