Question

Find zeros of the quadratic polynomial Root3x^2-8x+4root3
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Answer 1

Answer:

$\Large\sf\underline\red{Question:-} \\ \\ \sf \: find \: the \: root \: of \: quadratic \: \\ \sf \: equation \::- \sqrt{3} {x}^{2} - 8x + \sqrt{3} \\ \\ \\ \Large\sf\underline\red{answer} \\ \\ \sf\blue{ \: x = 2 \sqrt{3} \: \: \: , \: x = \frac{2}{ \sqrt{3} } } \\ \\ \\ \Large\sf\underline\red{Solution:-} \\ \\\sf \sqrt{3} {x}^{2} - 8x + \sqrt{3} = 0 \\ \\→\sf \sqrt{3} {x}^{2} - (6 + 2) + 4 \sqrt{3} = 0 \\ \\→ \: \: \: \: \sf \sqrt{3} {x}^{2} - 6x - x + \sqrt{3} = 0 \\ \\→ \sf \sqrt{3} x(x - 2 \sqrt{3} ) - 2(x - 2 \sqrt{3} ) =0 \\ \\→ \: \: \: \sf (x - 2 \sqrt{3} )( \sqrt{3} x - 2) = 0 \\ \\→ \: \: \: \: \sf x - 2 \sqrt{3} = 0 \\ \\\sf x = 2 \sqrt{3} \\ \\ \sf \: \blue{or }\\ \\→ \: \: \: \sf \sqrt{3} x - 2 = 0 \\ \\\sf x = \frac{2}{ \sqrt{3} }$