Question

Determine if the given quadratic equation has real roots and if so find the roots.
$\sqrt{3} {x}^{2} + 10x - 8 \sqrt{3} = 0$
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Answer 1

Answer:

GIVEN

root3x^2+10x-8root3=0

=root3x^2+12x-2x-8root3=0

=root3x(x+4root3)-2(x+4root3)=0

=(x+4root3)(root3x-2)=0

=x=-4root3(or) 2/root3

Answer 2

Answer:

Heya Mate

Here's your answer

$\sqrt{3} {x}^{2} + 10x - \sqrt[8]{3} = 0 \\ \sqrt{3} {x}^{2} + 12x - 2x - \sqrt[8]{3} = 0 \\ \sqrt{3} x (x + \sqrt[4]{3} ) - 2(x - \sqrt[4]{3} ) = 0 \\ ( \sqrt{3} x - 2)(x - \sqrt[4]{3} ) = 0 \\ ( \sqrt{3} x - 2) = 0 \\ \sqrt{3} x = 2 \\ x = 2 \times {3}^{2} \\ x = 18 \\( x - \sqrt[4]{3} ) = 0 \\ x = {3}^{4} \\ x = 81 \\ hence \: this \: equation \: have \: two \: roots \:$

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