Question

What are the roots of the quadratic equation root 3 x square - 2 x minus root 3 is equal to zero?

Answer 1

P(x) =√3x²-2x-√3

√3x²-3x+x-√3=0

√3x(x-√3) +1 ( x-√3)

(x-√3)(√3x+1)

x-3=0 √3x+1=0

x= 3 √3x= -1= x= -1/√3

Hence,

the roots of the equation are 3 and -1√3

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Answer 2

Given that,

A quadratic equation, $\sqrt{3} x^2-2x-\sqrt{3} =0$

To find,

The roots of the equation.

Solution,

We have,

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

We can solve it by using the middle term splitting method. So,

$\sqrt{3} x^2-2x-\sqrt{3} =0\\\\\sqrt{3} x^2 -3x+x-\sqrt3 =0$

Taking common terms,

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

It means,

$(x-\sqrt3)=0, (\sqrt3 x+1)=0\\\\x=\sqrt3, x=\dfrac{-1}{\sqrt3}$

Hence, two roots of the equation are $\sqrt{3}\ \text{and}\ \dfrac{-1}{\sqrt3}$.