Math, asked by BrainlyHelper, 1 year ago

Find the roots of the following quadratic equations (if they exist) by the method of completing the square. √2x²-3x-2√2=0

Answers

Answered by nikitasingh79
2

SOLUTION :  

Given : √2x² - 3x - 2√2 = 0

On dividing the whole equation by √2,

(x² - 3x/√2 - 2) = 0

Shift the constant term on RHS

x² - 3x/√2  =  2  

Add square of the ½ of the coefficient of x on both sides

On adding (½ of 3/√2)² = (3/2√2)² both sides

x² - 3x/√2 +  (3/2√2)² =  2 + (3/2√2)²

Write the LHS in the form of perfect square

(x - 3/2√2)² =  2 + 9/8

[a² - 2ab + b² = (a - b)²]

(x - 3/2√2)² = (2 × 8 + 9)/8

(x - 3/2√2)² = (16 + 9)/8

(x - 3/2√2)² = 25/8

On taking square root on both sides

(x - 3/2√2) = √(25/8)

(x - 3/2√2) = ± 5/√8

(x - 3/2√2) = ± 5/2√2

On shifting constant term (- 3/2√2) to RHS

x = ± 5/2√2 +  3/2√2

x =  (5 + 3)/2√2

[Taking + ve sign]

x = 8/2√2

x = 4/√2

On rationalising the denominator

x = (4 × √2) / √2 × √2

x = 4√2 /2

x = 2√2

x =  (- 5 + 3)/2√2

[Taking - ve sign]

x = -  2 /2√2

x = - 1/√2

Hence, the  roots of the given equation are  2√2 & - 1/√2 .

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Answered by mysticd
0

Solution :

Compare given Quadratic

equation √2x²-3x-2√2=0

with ax²+bx+c=0 , we get

a = √2 , b = -3 , c = -2√2

i ) Discreminant (D)

= b² - 4ac

= (-3)² - 4×(√2)(-2√2)

= 9 + 16

= 25

= 5²

D > 0

Therefore ,

Roots are real and distinct.

ii ) Finding roots by

completing square method:

√2x² - 3x - 2√2 = 0

Divide each term by √2,

we get

=> x² - (3/√2)x - 2 = 0

=> x² - 2•x•(3/2√2) = 2

=> x²-2•x•(3/2√2)+(3/2√2)²=2+(3/2√2)²

=> ( x - 3/2√2 )² = 2 + 9/8

=> ( x - 3/2√2 )² = (16 + 9 )/8

=> ( x - 3/2√2 )² = 25/8

=> x - 3/2√2 = ± √(25/8)

=> x = 3/2√2 ± 5/2√2

=> x = ( 3 ± 5 )/2√2

Therefore ,

x = (3+5)/2√2 Or x=(3-5)/2√2

=> x = 8/2√2 Or x = -2/2√2

=> x = 4/√2 Or x = -1/√2

=> x=(2•√2•√2)/√2 Or x = -1/√2

=> x = 2√2 Or x = -1/√2

••••

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