Please help me with the question provided in the attachment .
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Answers
Answer:
Given :
( x + 3 )( x + 4 )( x + 6 )( x + 7 ) = 1120
⇒ ( x + 3 )( x + 7 )( x + 4 )( x + 6 ) = 1120
⇒ ( x² + 3 x + 7 x + 21 )( x² + 4 x + 6 x + 24 ) = 1120
⇒ ( x² + 10 x + 21 )( x² + 10 x + 24 ) = 110
Let x² + 10 x be y .
⇒ ( y + 21 )( y + 24 ) = 1120
⇒ y² + 21 y + 24 y + 24×21 = 1120
⇒ y² + 45 y - 1120 + 504 = 0
⇒ y² + 45 y - 616 = 0
⇒ y² + 56 y - 11 y - 616 = 0
⇒ y ( y + 56 ) - 11 ( y + 56 ) = 0
⇒ ( y + 56 )( y - 11 ) = 0
Either y + 56 = 0 .
Or y - 11 = 0 .
When y + 56 = 0
x² + 10 x + 56 = 0
The quadratic equation will have no real roots because :
b² < 4 ac
⇒ 10² < 4×56
⇒ 100 < 224 .
Hence we neglect this value .
When y - 11 = 0
x² + 10 x - 11 = 0
⇒ x² + 11 x - x - 11 = 0
⇒ x ( x + 11 ) - 1 ( x + 11 ) = 0
⇒ ( x + 11 )( x - 1 ) = 0
Either x - 1 = 0
Or x + 11 = 0
Either x = 1 or x = - 11 .
Step-by-step explanation:
The given polynomial on the LHS of the equation is evidently an equation of degree 4 .
Hence substitution method is the quickest in this case .
If you are going for a competitive exam , then do it by trial and error method .
Simply taking out the factors of 1120 and then comparing both sides we can easily find the value of x .
Answer:
Given :
( x + 3 )( x + 4 )( x + 6 )( x + 7 ) = 1120
⇒ ( x + 3 )( x + 7 )( x + 4 )( x + 6 ) = 1120
⇒ ( x² + 3 x + 7 x + 21 )( x² + 4 x + 6 x + 24 ) = 1120
⇒ ( x² + 10 x + 21 )( x² + 10 x + 24 ) = 110
Let x² + 10 x be y .
⇒ ( y + 21 )( y + 24 ) = 1120
⇒ y² + 21 y + 24 y + 24×21 = 1120
⇒ y² + 45 y - 1120 + 504 = 0
⇒ y² + 45 y - 616 = 0
⇒ y² + 56 y - 11 y - 616 = 0
⇒ y ( y + 56 ) - 11 ( y + 56 ) = 0
⇒ ( y + 56 )( y - 11 ) = 0
Either y + 56 = 0 .
Or y - 11 = 0 .
When y + 56 = 0
x² + 10 x + 56 = 0
The quadratic equation will have no real roots because :
b² < 4 ac
⇒ 10² < 4×56
⇒ 100 < 224 .
Hence we neglect this value .
When y - 11 = 0
x² + 10 x - 11 = 0
⇒ x² + 11 x - x - 11 = 0
⇒ x ( x + 11 ) - 1 ( x + 11 ) = 0
⇒ ( x + 11 )( x - 1 ) = 0
Either x - 1 = 0
Or x + 11 = 0
Either x = 1 or x = - 11 .