Question

Find alpha and beta and verify
3x^2+2√3x+1

Answer 1

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

$\green{\tt{\therefore{\alpha=\frac{-\sqrt{3}}{3}}}}$

$\green{\tt{\therefore{\beta=\frac{-\sqrt{3}}{3}}}}$

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

$\green{\underline \bold{Given:}} \\ \tt: \implies 3{x}^{2} + 2 \sqrt{3} x + 1 = 0 \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Value \: of \: \alpha \: and \: \beta = ?$

• According to given question :

$\tt: \implies 3 {x}^{2} + 2 \sqrt{3}x + 1 = 0 \\ \\ \tt\circ \: a = 3 \\ \\ \tt \circ \: b = 2 \sqrt{3} \\ \\ \tt \circ \: c = 1 \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies D = {b}^{2} - 4ac \\ \\ \tt: \implies D= {(2 \sqrt{3} )}^{2} - 4 \times 3 \times 1 \\ \\ \tt: \implies D =12 - 12 \\ \\ \tt: \implies D=0 \\ \\ \tt: \implies x = \frac{ - b \pm \sqrt{d} }{2a} \\ \\ \tt: \implies d = \frac{ - 2 \sqrt{3} \pm \sqrt{0} }{2 \times 3} \\ \\ \tt: \implies x = \frac{ - 2 \sqrt{3} }{6} \\ \\ \green{\tt: \implies x = \frac{ - \sqrt{3} }{3} } \\ \\ \green{\tt\therefore value \: of \: \alpha = \frac{ - \sqrt{3} }{3} } \\ \\ \green{\tt\therefore value \: of \: \beta = \frac{ - \sqrt{3} }{3} }$

Answer 2

$</p><p>\tt{\huge{\pink {Hello!!! }}}$

$</p><p>\tt{\green{Given \: Quadratic \: Equation \: - }}$

$\tt{ \red { = > 3 {x}^{2} + 2 \sqrt{3} x + 1}}$

$\tt{ \blue{d \: = \: \sqrt{ {b}^{2} - 4ac \: } = 0 } }</p><p>$

$</p><p>\tt{\orange{Values \: Of \: \alpha \: and \: \beta = \begin{cases}</p><p></p><p>\dfrac{1}{\sqrt{3}} & \text{alpha} \\</p><p> \\ \\</p><p>\dfrac{-1}{\sqrt{3}} & \text{beta}</p><p></p><p>\end{cases}}}</p><p>$

$</p><p>\tt{\huge{\purple {Verification \: - }}}$

We Know That The Value Of D is zero.

Hence D^2 is zero.

=>b^2 = 4 ac.

Now verify.