Question

Solve the Quadratic equation by Factorization.

$i) \: 4x {}^{2} - 4a {}^{2} x + (a {}^{4} - b {}^{4} ) = 0 \\ \\ ii) \: 9x {}^{2} - 9(a + b)x + (2a {}^{2} + 5ab + 2b {}^{2} ) = 0$

Answer 1

Hey there !!

▶ Question 1 :-

4x² - 4a²x + ( a⁴ - b⁴ ) = 0.

We write, -4a²x = -2( a² + b² )x - 2( a² - b² )x .

Because, 4x² × ( a⁴ - b⁴ ) = 4( a⁴ - b⁴ )x².

= [ -2( a² + b² )]x × [ -2( a² - b² )]x .

•°• 4x² - 4a²x + ( a⁴ - b⁴ ) = 0.

=> 4x² -2( a² + b² )x -2( a² - b² )x + ( a² - b² ) ( a² + b² ) = 0.

=> 2x[ 2x - ( a² + b² )] - ( a² - b² )[ 2x - ( a² + b² )] = 0.

=> [ 2x - ( a² + b² ) ] [ 2x - ( a² - b² ) ] = 0.

=> 2x - ( a² + b² ) = 0 or 2x - ( a² - b² ) = 0.

•°• x = ( a² + b² )/2 or x = ( a² - b² )/2 .

▶ Question 2 :-

9x² - 9( a + b )x + ( 2a² + 5ab + 2b² ) = 0.

We write, -9( a + b )x = -3( 2a + b )x - 3( a +2b )x.

Because, 9x² × ( 2a² + 5ab + 2b² ) = 9( 2a² + 5ab + 2b² )x² .

= [ -3 ( 2a + b )x ] × [ -3( a + 2b )x ] .

•°• 9x² - 9( a + b )x + ( 2a² + 5ab + 2b² ) = 0.

=> 9x² -3( 2a + b )x - 3( a + 3b )x + ( 2a + b )( a + 2b ) = 0.

=> 3x[ 3x - ( 2a + b ) ] - ( a + 2b) [ 3x - ( 2a + b ) ] = 0.

=> [ 3x - ( 2a + b ) ] [ 3x - ( a + 2b ) ] = 0.

=> 3x - ( 2a + b ) = 0 or 3x - ( a + 2b ) = 0.

=> x = $\frac{ ( 2a + b )}{3}$ or x = $\frac{( a + 2b )}{3} .$

✔✔ Hence, it is solved ✅✅.

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