If (x + a) is a common factor of x2 –

Solution

x² – (2+√2)x + 2√2 = 0 ------ (i) x² - 2√2x + 2 = 0 ------ (ii) Now, x + a = 0 => x = -a ------ (iii) Substituting in (i) and (ii), => x² - (2 + √ 2) x + 2√2 = 0 => a² - ( 2 + √ 2 ) (-a) + 2√ 2 = 0 => a² + ( 2 + √ 2 ) a + 2√ 2 = 0 => x² - 2√ 2 x + 2 = 0 => a² + 2√ 2 a + 2 = 0 Now, => a² + ( 2 + √ 2 ) a + 2√ 2 = a² + 2√ 2 a + 2 2a + √ 2 a + 2√ 2 = 2√ 2 a + 2 2a + √ 2a - 2√ 2 a = 2 - 2√ 2 2a - √ 2a = 2 - 2√ 2 a ( 2 - √ 2 ) =  2 - 2√ 2 a = ( 2 - 2√ 2 ) / ( 2 - √ 2 ) Substituting the value of a in the equation: √ ( 5√ 2 - a ) 3√ 2 = √ ( 5√ 2 - [ ( 2 - 2√ 2 ) / ( 2 - √ 2 ) ] × 3√ 2 = √ [ ( 5√ 2 × ( 2 - √ 2 ) ) - ( 2 - 2√ 2 )  /  ( 2 - √ 2 ) ] × 3√ 2 = √ [ ( 10√ 2 - 10 - 2 + 2√ 2 ) / ( 2 - √ 2 ) ] × 3√ 2 = √ [ ( 12√ 2 - 12 ) / ( 2 - √ 2 ) ] × 3√ 2 = √ [ ( (36 × 2) - ( 36 × √ 2 ) ) / ( 2 - √ 2 ) ] = √ ( 72 - 36√ 2) / ( 2 - √ 2 ) = √ 36 ( 2 - √ 2 ) / ( 2 - √ 2 ) = √ 36 = 6 ∴ √ ( 5√ 2 - a ) 3√ 2 = 6