Question

of x=7+4√3 then find the value of x + 1 / 2​

Answer 1

Step-by-step explanation:

The value of x+\frac{1}{x}x+x1 = 1414

Explanation:

Given x = 7-4√3 ----(1)

i) \frac{1}{x}x1

=\frac{1}{7-4\sqrt{3}}7−431

Rationalising the denominator, we get

= \frac{7+4\sqrt{3}}{(7-4\sqrt{3})(7+4\sqrt{3})}(7−43)(7+43)7+43

= \frac{7+4\sqrt{3}}{7^{2}-\left(4\sqrt{3}\right)^{2}}72−(43)27+43

/* By algebraic identity

(a+b)(a-b) = a² - b² */

= \frac{7+4\sqrt{3}}{(49-48)}(49−48)7+43

= 7+4\sqrt{3}7+43 ---(2)

Now ,

The value of x+\frac{1}{x}x+x1

= 7-4\sqrt{3}+7+4\sqrt{3}7−43+7+43

= 1414

Therefore,.

The value of x+\frac{1}{x}x+x1 = 1414

Answer 2

Answer: 15 - (8 under root 3) /2.

Step-by-step explanation:

x = 7- 4(under root 3)

according to question,

x + 1/2

= 7- 4(under root 3) + 1/2

= 2[ 7- 4(under root 3)] /1 +1/2

= [14 - (8 root 3) + 1] /2

= 15 - (8 under root 3) /2.

I HOPE IT HELPS