Question

Solve for x: x^2-(root3+1)x+root3=0

Answer 1

required answer is here...

Answer 2

Answer:

$x=\sqrt{3},1$

Step-by-step explanation:

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

To find : Solve for x?

Solution :

The given expression is in quadratic form,

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

Open the bracket,

$x^2-\sqrt{3}x -x+\sqrt{3} =0$

Arrange in order s.t. there is common,

$x(x-\sqrt{3}) -1(x-\sqrt{3})=0$

$(x-\sqrt{3})(x-1)=0$

$(x-\sqrt{3})=0,(x-1)=0$

$x=\sqrt{3},x=1$

Therefore, The value of x is $x=\sqrt{3},1$