Question

If (7-4root 3)x^2+4x+3+(7+4 root 3)x^2-4x+3=14,then the value of x is given by​

Answer 1

Given:

           $(7 + 4\sqrt{3} )^{x^2 - 4x + 3} + (7 - 4\sqrt{3} )^{x^2 - 4x + 3} = 14$

To Find:

           The value of $x$

Solution:

Let

      $(7 + 4\sqrt{3} )^{x^2 - 4x + 3} = y.$

Then,

The given equation becomes,

⇒     $y + \frac{1}{y} = 14$

⇒    $y^2 - 14x + 1 = 0$

⇒    $y = \sqrt{ 4 +_- \sqrt{(196 - 4)}}$

⇒   $y = 7 +_-4\sqrt{3}$

⇒   $(7 + 4\sqrt{3} )^{x2 - 4x + 3} = 7 +_- 4\sqrt{3}$

⇒    $(7 + 4\sqrt{3} )^{x2 - 4x + 3} = (7 +_- 4\sqrt{3})^1$

⇒ $(7 + 4\sqrt{3} )^{x^2 + 4x + 3} = (7 - 4\sqrt{3} ) = (7 + 4\sqrt{3} )^{-1}$

since,

           $7 - 4\sqrt{3} = \frac{1}{7 + 4\sqrt{3}}$

⇒     $x^2 - 4x + 3 = 1 \ or\ x^2 - 4x + 3 = -1$

⇒    $x^2 - 4x + 2 = 0 \or, \ x^2 - 4x + 4 = 0$

⇒   $x = \frac{4 +_- \sqrt{16-8}}{2} \ or \ (x - 2)^2 = 0$

⇒  $x = 2+_- \sqrt{2} , 2$

Hence, the roots of the given equation are $x = 2+_- \sqrt{2} , 2$

Answer 2

Answer:

Value of x can be

$2 + \sqrt{2}$

$2 - \sqrt{2}$

and

$2$