Question

$(x {}^{2} + 2x)(x {}^{2} + 2x - 11) + 24 = 0$
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Answer 1

Answer:

Possible values of x are - 4 , 2 , - 3 and 1.

Step-by-step explanation:

Given,

( x^2 + 2x )( x^2 + 2x - 11 ) + 24 = 0

Let : x^2 + 2x = a

Continued :

= > a( a - 11 ) + 24 = 0

= > a^2 - 11a + 24 = 0

= > a^2 - ( 8 + 3 )a + 24 = 0

= > a^2 - 8a - 3a + 24 = 0

= > a( a - 8 ) - 3( a - 8 ) = 0

= > ( a - 8 )( a - 3 ) = 0

= > a = 8 or 3

Case 1 : a = 8

= > x^2 + 2x = 8

= > x^2 + 2x - 8 = 0

= > x^2 + ( 4 - 2 )x - 8 = 0

= > x^2 + 4x - 2x - 8 = 0

= > x( x + 4 ) - 2( x + 4 ) = 0

= > ( x + 4 )( x - 2 ) = 0

= > x = - 4 or x = 2

Case 2 :

= > x^2 + 2x = 3

= > x^2 + 2x - 3 = 0

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

= > x^2 + 3x - x - 3 = 0

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

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

= > x = - 3 or 1

Possible values of x are - 4 , 2 , - 3 and 1.

Answer 2

Question :- Solve (x²+2x)(x² + 2x - 11) + 24 = 0 .

Solution :-

Let (x²+2x) = y .

Putting y in Question now, we get,

→ y² - 11y + 24 = 0

Splitting the Middle Term now,

→ y² - 8y - 3y + 24 = 0

→ y ( y - 8 ) - 3( y - 8) = 0

→ ( y - 8 ) ( y - 3 ) = 0

Putting Again Back y as (x² + 2x) now ,

→ ( x² + 2x - 8 ) ( x² + 2x - 3 ) = 0

Splitting the Middle term now,

→ ( x² + 4x - 2x - 8 ) ( x² + 3x - 1x - 3) = 0

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

→ [(x+4)(x-2)] * [(x+3)(x-1)] = 0

Now, putting All Equal to 0 , we get,

→ x + 4 = 0 , =>> x = (-4)

→ x - 2 = 0 , =>> x = 2

→ x + 3 = 0 , =>> x = (-3)

→ x - 1 = 0 , =>> x = 1

Hence, value of x will be (-4), 2 , (-3) and 1 .