Question

if 2 x square equal to 2 + root 3 write X in terms of a + b root 3​

Answer 1

This is form of x in terms of 2 + $\sqrt{3}$ is "1 + $\dfrac{1}{2}$·$\sqrt{3}$".

Step-by-step explanation:

Given,

2$x^{2}$ = 2 + $\sqrt{3}$

⇒ $x^{2}$  = $\dfrac{2}{2}$ + $\dfrac{[tex]\sqrt{3}$ }{2}[/tex]

⇒ $x^{2}$ = 1 + $\dfrac{1}{2}$·$\sqrt{3}$

This is form of x in terms of 2 + $\sqrt{3}$ is 1 + $\dfrac{1}{2}$·$\sqrt{3}$ .

Answer 2

$x=\dfrac{1}{2}+\dfrac{1}{2}\sqrt{3}$

or

$x=-\dfrac{1}{2}-\dfrac{1}{2}\sqrt{3}$

Step-by-step explanation:

Given that, $2x^2=2+\sqrt{3}$

$x^2=1+\dfrac{1}{2}\sqrt{3}$

$x=\sqrt{1+\dfrac{1}{2}\sqrt{3}}$

Let $\sqrt{1+\dfrac{1}{2}\sqrt{3}}=a+b\sqrt{3}$

where, a and b are real number.

Taking square both sides

$1+\dfrac{1}{2}\sqrt{3}=(a+b\sqrt{3})^2$

$1+\dfrac{1}{2}\sqrt{3}=a^2+3b^2+2ab\sqrt{3}$

Comparing the coefficients

$a^2+3b^2=1$ ------- (1)

$2ab=\dfrac{1}{2}$

$b=\dfrac{1}{4a}$

Substitute b into eq(1)

$a^2+\dfrac{3}{16a^2}=1$

$16a^4-16a^2+3=0$

using quadratic formula, $ax^2+bx+c=0$ then $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

Therefore,

$a^2=\dfrac{16\pm\sqrt{16^2-4\cdot 16\cdot 3}}{2\cdot 16}$

$a^2=\dfrac{3}{4},\dfrac{1}{4}$

$a=\pm\dfrac{\sqrt{3}}{2},\pm\dfrac{1}{2}$

a is real. So, possible value of, $a=\pm\dfrac{1}{2}$

Put a into $b=\dfrac{1}{4a}$

$b=\pm\dfrac{1}{2}$

So, $x=\dfrac{1}{2}+\dfrac{1}{2}\sqrt{3}$

or $x=-\dfrac{1}{2}-\dfrac{1}{2}\sqrt{3}$

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