SOLUTION: Find the REAL Solution of the equation: 2(x+1)^2+14(x+1)+20=0

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Question 174541: Find the REAL Solution of the equation:
2(x+1)^2+14(x+1)+20=0
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

2%28x%2B1%29%5E2%2B14%28x%2B1%29%2B20=0 Start with the given equation.


Let z=x%2B1


2z%5E2%2B14z%2B20=0 Replace each x%2B1 with "z"


Notice we have a quadratic equation in the form of az%5E2%2Bbz%2Bc where a=2, b=14, and c=20


Let's use the quadratic formula to solve for z


z+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29 Start with the quadratic formula


z+=+%28-%2814%29+%2B-+sqrt%28+%2814%29%5E2-4%282%29%2820%29+%29%29%2F%282%282%29%29 Plug in a=2, b=14, and c=20


z+=+%28-14+%2B-+sqrt%28+196-4%282%29%2820%29+%29%29%2F%282%282%29%29 Square 14 to get 196.


z+=+%28-14+%2B-+sqrt%28+196-160+%29%29%2F%282%282%29%29 Multiply 4%282%29%2820%29 to get 160


z+=+%28-14+%2B-+sqrt%28+36+%29%29%2F%282%282%29%29 Subtract 160 from 196 to get 36


z+=+%28-14+%2B-+sqrt%28+36+%29%29%2F%284%29 Multiply 2 and 2 to get 4.


z+=+%28-14+%2B-+6%29%2F%284%29 Take the square root of 36 to get 6.


z+=+%28-14+%2B+6%29%2F%284%29 or z+=+%28-14+-+6%29%2F%284%29 Break up the expression.


z+=+%28-8%29%2F%284%29 or z+=++%28-20%29%2F%284%29 Combine like terms.


z+=+-2 or z+=+-5 Simplify.


x%2B1+=+-2 or x%2B1+=+-5 Plug in z=x%2B1 for each equation.


x+=+-2-1 or x+=+-5-1 Subtract 1 from both sides (for each equation).


x+=+-3 or x+=+-6 Combine like terms (for each equation).


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Answer:


So the solutions are x+=+-3 or x+=+-6


Note: both of these solutions are real.