Question

solve the following quadratic equation by factorisation
3x^2-2√6x+2=0​

Answer 1

Answer:

3x²-2√6x+2=0

a²-2ab+b²=(a-b)²

a=√3x      b=√2      2ab= 2×√3x×√2=2√6x

3x²-2√6x+2=(√3x-√2)²

(√3x-√2)(√3x-√2)=0

√3x-√2=0

√3x=√2

x=√2/√3

Answer 2

GIVEN :-

• A quadratic equation = 3x² - 2√6x + 2 = 0.

TO FIND :-

• Zeros of given quadratic equation.

SOLUTION :-

$\\ : \implies \displaystyle \sf \: 3x ^{2} - 2\sqrt{6} x + 2 = 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \tiny{ \dag \displaystyle \sf \:By \: splitting \: the \: middle \: term \: of \: quadratic \: equation.}$

$: \implies \displaystyle \sf 3x ^{2} - \sqrt{6} x - \sqrt{6} x + 2 = 0$

Here we can write , 3x² as (√3x)(√3x) and √6x as (√3x)(-√2) and 2 as √2 × √2.

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

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

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

$: \implies \displaystyle \sf \sqrt{3}x = 0 + \sqrt{2} \: , \: \sqrt{3} x = 0 + 2$

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

$: \implies \underline{ \boxed{ \displaystyle \sf \: x = \frac{ \sqrt{3} }{ \sqrt{2} } \: , \: x = \frac{ \sqrt{3} }{ \sqrt{2} } }}$