Tag: solving a quadratic equation

Questions Related to solving a quadratic equation

Solve the following quadratic equation by completing the square: $ x^2+(\sqrt{3}+1)x+\sqrt{3}=0$

maths factorization-1 solution of a quadratic equation by completing the square solving a quadratic equation solving a quadratic equation: completing the square

  1. $\left { \sqrt{3}, 1\right }$

  2. $\left { \sqrt{3}, 2\right }$

  3. $\left { \sqrt{3}, 3\right }$

  4. None of these

Show answer
Correct Option: A
Explanation:

Given equation is $x^2-(\sqrt{3}+1)x+\sqrt{3}=0$


$\Rightarrow$  $x^2-(\sqrt{3}+1)x=-\sqrt{3}$


Now, adding $\left(\dfrac{\sqrt{3}+1}{2}\right)^2$ on both sides,

$\Rightarrow$  $x^2-2\times x\times \left(\dfrac{\sqrt{3}+1}{2}\right)+\left(\dfrac{\sqrt{3}+1}{2}\right)^2=-\sqrt{3}+\left(\dfrac{\sqrt{3}+1}{2}\right)^2$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{(\sqrt{3}+1)^2}{4}-\sqrt{3}\right]$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{3+2\sqrt{3}+1-4\sqrt{3}}{4}\right]$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{3-2\sqrt{3}+1}{4}\right]$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{\sqrt{3}-1}{2}\right]^2$

Taking square root on both sides, 
  
$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=\pm\dfrac{\sqrt{3}-1}{2}$

$\Rightarrow$  $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=\dfrac{\sqrt{3}-1}{2}$ 
and $\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=-\dfrac{\sqrt{3}-1}{2}$

$\therefore$  $x=\dfrac{2\sqrt{3}}{2}$  and $ x=\dfrac{2}{2}$

$\therefore$  $x=\sqrt{3}$ and $x=1$