Question

(i) 3x2 - 4 V3 x+4 = 0​

Answer 1

GIVEN :-

• 3x² - 4√3x + 4 = 0.

TO FIND :-

• The value of x.

SOLUTION :-

$\implies \sf \: 3x ^{2} - 4 \sqrt{3} x + 4 = 0$

By splitting the middle term,

$\implies \sf \: 3x ^{2} - \bigg (2 \sqrt{3} + 2 \sqrt{3} \bigg )x + 4 = 0$

$\implies \sf \: 3x ^{2} - 2 \sqrt{3}x - 2 \sqrt{3}x + 4 = 0$

$\implies \sf \: \sqrt{3} x \bigg( \sqrt{3} x - 2 \bigg) - 2 \bigg( \sqrt{3} x - 2 \bigg) = 0$

$\implies \sf \: \bigg( \sqrt{3} x - 2 \bigg) \bigg( \sqrt{3} x - 2 \bigg) = 0$

$\implies \sf \: \bigg( \sqrt{3} x - 2 \bigg) ^{2} = 0$

$\implies \sf \: \sqrt{3}x - 2 = \sqrt{0}$

$\implies \sf \: \sqrt{3} x - 2 = 0$

$\implies \sf \: \sqrt{3} x = 2$

$\implies \underline{ \boxed{ \sf \: x = \dfrac{2}{ \sqrt{3} } }}$

The above quadratic equation has real and equal roots .

Answer 2

✰TO SOLVE.

• $\\ \tt{3x {}^{2} - 4 \sqrt{3} x + 4 = 0}$

✰EXPLANATION.

➣$\tt3 { x }^{ 2 } -4 \sqrt{ 3 } x+4 = 0$

➣$\tt \: 3x^{2}+\left(-4\sqrt{3}\right)x+4=0$

➣$\tt \: x=\frac{-\left(-4\sqrt{3}\right)±\sqrt{\left(-4\sqrt{3}\right)^{2}-4\times 3\times 4}}{2\times 3}$

➣$\tt \: x=\frac{-\left(-4\sqrt{3}\right)±\sqrt{48-4\times 3\times 4}}{2\times 3}$

➣$\tt \: x=\frac{-\left(-4\sqrt{3}\right)±\sqrt{48-12\times 4}}{2\times 3}$

➣$\tt \: x=\frac{-\left(-4\sqrt{3}\right)±\sqrt{48-48}}{2\times 3}$

➣$\tt \: x=\frac{-\left(-4\sqrt{3}\right)±\sqrt{0}}{2\times 3}$

➣$\tt \: x=-\frac{-4\sqrt{3}}{2\times 3}$

➣$\tt\: x=\frac{4\sqrt{3}}{2\times 3}$

➣$\tt \: x=\frac{4\sqrt{3}}{6}$

➣$\tt \: x=\frac{2\sqrt{3}}{3}$

The Quadric Equation has real and equal roots.

✰ Hope it helps u ✰