Question

solve x^2-(√3+1)x+√3=0 by completing square

Answer 1

x²-(3^½ + 1)x + 3^½ = 0
(x - 3^½)(x - 1) = 0
x = 3^½     or,     x=1,


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Answer 2

Answer:

The solution of the given equation is $x=\sqrt3,x=1$

Step-by-step explanation:

Given : Expression  $x^2-(\sqrt{3}+1)x+\sqrt3$

To find : Solve by completing square?

Solution :  

The general form of quadratic equation is $ax^2+bx+c=0$

To convert into complete square the form is $a(x+d)^2+e=0$

Where, $d=\frac{b}{2a} \text{ and } e=c-\frac{b^2}{4a}$

Now, comparing the given equation $x^2-(\sqrt{3}+1)x+\sqrt3$

a=1 , $b=-\sqrt{3}+1$, $c=\sqrt{3}$    

$d=\frac{b}{2a}=-\frac{\sqrt3+1}{2(1)}=-\frac{\sqrt{3}+1}{2}$

$e=c-\frac{b^2}{4a}=\sqrt3-\frac{(\sqrt3+1)^2}{4(1)}=\frac{\sqrt3+1}{2}$

Substitute in $a(x+d)^2+e=0$

$1(x-\frac{\sqrt{3}+1}{2})^2-\frac{\sqrt3-1}{2}=0$

$(x-\frac{\sqrt{3}+1}{2})^2-\frac{\sqrt3-1}{2}=0$

$(x-\frac{\sqrt{3}+1}{2})^2=\frac{\sqrt3-1}{2}$

Root both side,

$x-\frac{\sqrt{3}+1}{2}=\pm\frac{\sqrt3-1}{2}$

$x-\frac{\sqrt{3}+1}{2}=\frac{\sqrt3-1}{2}$ or $x-\frac{\sqrt{3}+1}{2}=-\frac{\sqrt3-1}{2}$

When we solve we get,

$x=\sqrt3,x=1$

Therefore, The solution of the given equation is $x=\sqrt3,x=1$