Question

solve the following quadratic equation by factorization. x2 -(√3+1)x+√3=0​

Answer 1

Answer:

$x = 1 \: or \: \sqrt{3}$

Step-by-step explanation:

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

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

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

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

$\mathsf{x = 1 \: or \: \sqrt{3}}$

Answer 2

Answer:

Answer:

x = 1 \: or \: \sqrt{3}x=1or

3

Step-by-step explanation:

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

2

−(

3

+1)x+

3

=0

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

2

−

3x

−x+

3

=0

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

3

)−1(x−

3

)=0

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

3

)=0

\mathsf{x = 1 \: or \: \sqrt{3}}x=1or

3