Question

Solve the following quadratic equations by factorization: √3x²-2√2x-2√3=0

Answer 1

SOLUTION :  

Given : √3x² - 2√2x - 2√3 = 0

√3x² - 3√2x + √2x - 2√3 = 0

[ 3√2× √2 = 6 & -3√2 + √2= - 2√2]

√3x(x - √6) + √2(x - √6) = 0

(√3x +√2) (x - √6) = 0

√3x +√2 = 0   or  x - √6 = 0  

√3x = - √2  or  x = √6

x =  - √2/√3  or  x = √6

x =  - √(⅔)  or  x = √6

Hence, the roots of the quadratic equation √3x² - 2√2x - 2√3 = 0 are - √(⅔) &  -√6 .

★★ METHOD TO FIND SOLUTION OF a quadratic equation by FACTORIZATION METHOD :  

We first write the given quadratic polynomial as product of two linear factors by splitting the middle term and then equate each factor to zero to get desired roots of given quadratic equation.

HOPE THIS ANSWER WILL HELP YOU….

Answer 2

√3x² - 2√2x - 2√3 = 0

Splitting the middle term of the given quadratic equation,

=> √3x² - 3√2x + √2x - 2√3 = 0

Taking out common factors,

=> √3x * ( x - √6 ) + √2 * ( x - √6 ) = 0

Again taking out common factors,

=> ( x - √6 ) * ( √3x + √2 ) = 0

We can omit writing " * " between the given terms in brackets so the actual result,

=> ( x - √6 ) ( √3x + √2 ) = 0

Therefore following the zero product rule,

=> ( x - √6 ) = 0 OR ( √3x + √2 ) = 0
=> x = √6 OR x = - √2 / √3
=> x = √6 OR x = - √(⅔)

Therefore, we conclude that, the roots of the given quadratic equation are,
√6 and -√(⅔)