Question

show that x=√3 is a solution of equation x^2-3√3+6= 0​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Value\:of\:x=\sqrt{3}\:and\:2\sqrt{3}}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies {x}^{2} - 3 \sqrt{3} x + 6 = 0 \\ \\ \red{\underline \bold{To \: Show:}} \\ \tt: \implies x = \sqrt{3}$

• According to given question :

$\tt \circ \: {x}^{2} - 3 \sqrt{3x} + 6 = 0 \\ \\ \tt \circ \: a = 1 \\ \\ \tt \circ \: b = - 3 \sqrt{3} \\ \\ \tt \circ \: c = 6 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies D = {b}^{2} - 4ac \\ \\ \tt: \implies D = { (- 3 \sqrt{3} )}^{2} - 4 \times 1 \times 6 \\ \\ \tt: \implies D = 27 - 24 \\ \\ \green{\tt: \implies D = 3 } \\ \\ \tt: \implies x= \frac{ - b \pm \sqrt{D} }{2a} \\ \\ \tt: \implies x = \frac{ - ( - 3 \sqrt{3}) \pm \sqrt{3} }{2 \times 1} \\ \\ \tt: \implies x= \frac{3 \sqrt{3} \pm \sqrt{3} }{2} \\ \\ \tt: \implies x = \frac{3 \sqrt{3} + \sqrt{3} }{2} \: and \: \frac{3 \sqrt{3} - \sqrt{3} }{2} \\ \\ \tt: \implies x = \frac{4 \sqrt{3} }{2} \: and \: \frac{2 \sqrt{3} }{2} \\ \\ \green{\tt: \implies x= 2 \sqrt{3} \: and \: \sqrt{3} }$