Question

find the quadratic polynomial
$\sqrt{3} \times {x}^{2} - 8x + 4 \sqrt{3}$
​

Answer 1

Question :-

Solve this Quadratic polynomial -

$\to \sf \sqrt{3} {x}^{2} - 8x + 4 = 0$

Answer :-

$\to \boxed{\sf \: x = 2 \sqrt{3 \: } \: or \: \frac{2}{ \sqrt{3} } }$

Used formula :-

Here we will use Shridharacharya principle -

if any quadratic equation f(x) = ax² + bx + c = 0

,

$\to \sf \: x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

Solution :-

We have ,

$\to \sf \: \sqrt{3} {x}^{2} - 8x + 4 \sqrt{3} = 0 \\ \sf comparing \: wit h \: \\ \to \sf \: a{x}^{2} + bx + c = 0\\ \\ \to \sf \: a = \sqrt{3} , b = - 8 \: and \: c = 4 \sqrt{3} \\ \\ \sf \: hence \\ \: \\ \to \sf \: x = \frac{ - ( - 8) \pm \sqrt{ {( - 8)}^{2} - 4 \sqrt{3} \times 4 \sqrt{3} } }{2 \sqrt{3} } \\ \\ \to \sf \: x = \frac{8 \pm \sqrt{64 - 48} }{2 \sqrt{3} } \\ \\ \to \sf \: x = \frac{8 \pm \: \sqrt{16} }{2 \sqrt{3} } \\ \\ \to \sf \: x = \frac{8 \pm \: 4}{2 \sqrt{3} } \\ \\ \to \sf \: x = \frac{4 \pm \: 2}{ \sqrt{3} } \\ \\ \star \sf \: case \: (1) \: if \: \\ \\ \to \sf \: x = \frac{4 - 2}{ \sqrt{3} } \\ \\ \to \: \boxed{\sf x = \frac{2 }{ \sqrt{3} } } \\ \\ \star \sf \: case \: (2) \: if \\ \\ \to \sf \: x = \frac{4 + 2}{ \sqrt{3} } \\ \\ \to \sf \: x = \frac{6}{ \sqrt{3} } \\ \\ \to \: \: \boxed{ \sf x = 2 \sqrt{3} }$

Verification :-

$(1) \sf \: when \:x = 2 \sqrt{3} \\ \\ \to \: \sqrt{3} {(2 \sqrt{3}) }^{2} - 8 \times 2\sqrt{3} + 4 \sqrt{3} = 0 \\ \\ \to \:12 \sqrt{3} - 16 \sqrt{3} + 4 \sqrt{3} = 0 \\ \\ \to \: 16 \sqrt{3} - 16 \sqrt{3} = 0 \\ \\ \to \: 0 = 0 \\ \\ (2) \sf \: when \: x = \frac{2}{ \sqrt{3} } \\ \\ \to \: \sqrt{3} {( \frac{2}{ \sqrt{3} }) }^{2} - 8 \times \frac{2}{ \sqrt{3} } + 4 \sqrt{3} = 0 \\ \\ \to \: \frac{4 \sqrt{3} }{3} - \frac{16}{ \sqrt{3} } + 4 \sqrt{3} = 0 \\ \\ \to \: \frac{4 \sqrt{3} \sqrt{3} - 48 + 4 \sqrt{3} \times 3 \sqrt{3} }{3 \sqrt{3} } = 0 \\ \\ \to \: \frac{12 - 48 + 36}{3 \sqrt{3} } = 0 \\ \\ \to \: \frac{48 - 48}{3 \sqrt{3} } = 0 \\ \\ \to \: 0 = 0 \\ \\ \large \sf hence \: verified$