Question

Solve the following quadratic equations by factorization method.

Answer 1

Hey user

Here is your answer :-

Let ( x + 1 ) = a

2a^2 - 5a = 12

》2a^2 - 5a - 12 = 0

》2a^2 - 8a + 3a - 12 = 0

》2a ( a - 4 ) + 3 ( a - 4 ) = 0

》( a - 4 )( 2a + 3 ) = 0

》( a - 4 ) = 0. or. ( 2a + 3 ) = 0

》 a = 4. or. a = - 3/2

》( x + 1 ) = 4. or. ( x + 1 ) = -3/2
》x = 3. or. x = - 5/2

Thank you .

Answer 2

Given Quadratic equation ,

2(x+1)²-5(x+1)= 12

=> 2(x+1)²-5(x+1)-12= 0

Splitting the middle term ,

=>2(x+1)²-8(x+1)+3(x+1)-12=0

=> 2(x+1)[x+1-4] + 3[ x+1 - 4] = 0

=> 2( x + 1 )( x - 3) + 3( x - 3 ) = 0

=> ( x - 3 ) [ 2( x + 1 ) + 3 ] = 0

=> ( x - 3 ) ( 2x + 2 + 3 ) = 0

=> ( x - 3 ) ( 2x + 5 ) = 0

x - 3 = 0 or 2x + 5 = 0

x = 3 or x = -5/2

Therefore ,

x = 3 or x = -5/2

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