The solutions to x 2 + 6 x + 7 = 0 are r and s , where r <

Question

The solutions to x 2 + 6 x + 7 = 0 are r and s , where r &#60; s . The solutions to x 2 + 8 x + 8 = 0 are t and u , where t &#60; u . The solutions to x 2 + 14 x + c = 0 , where c is a constant, are r + t and s + u . What is the value of c ?

Accepted Answer

The correct answer is 31 . Subtracting 7 from both sides of the equation x 2 + 6 x + 7 = 0 yields x 2 + 6 x = - 7 . To complete the square, adding 6 2 2 , or 3 2 , to both sides of this equation yields x 2 + 6 x + 3 2 = - 7 + 3 2 , or x + 3 2 = 2 . Taking the square root of both sides of this equation yields x + 3 = &#177; 2 . Subtracting 3 from both sides of this equation yields x = - 3 &#177; 2 . Therefore, the solutions r and s to the equation x 2 + 6 x + 7 = 0 are - 3 - 2 and - 3 + 2 . Since r &#60; s , it follows that r = - 3 - 2 and s = - 3 + 2 . Subtracting 8 from both sides of the equation x 2 + 8 x + 8 = 0 yields x 2 + 8 x = - 8 . To complete the square, adding 8 2 2 , or 4 2 , to both sides of this equation yields x 2 + 8 x + 4 2 = - 8 + 4 2 , or x + 4 2 = 8 . Taking the square root of both sides of this equation yields x + 4 = &#177; 8 , or x + 4 = &#177; 2 2 . Subtracting 4 from both sides of this equation yields x = - 4 &#177; 2 2 . Therefore, the solutions t and u to the equation x 2 + 8 x + 8 = 0 are - 4 - 2 2 and - 4 + 2 2 . Since t &#60; u , it follows that t = - 4 - 2 2 and u = - 4 + 2 2 . It's given that the solutions to x 2 + 14 x + c = 0 , where c is a constant, are r + t and s + u . It follows that this equation can be written as x - r + t x - s + u = 0 , which is equivalent to x 2 - r + t + s + u x + r + t s + u = 0 . Therefore, the value of c is r + t s + u . Substituting - 3 - 2 for r , - 4 - 2 2 for t , - 3 + 2 for s , and - 4 + 2 2 for u in this equation yields - 3 - 2 + - 4 - 2 2 - 3 + 2 + - 4 + 2 2 , which is equivalent to - 7 - 3 2 - 7 + 3 2 , or - 7 - 7 - 3 2 3 2 , which is equivalent to 49 - 18 , or 31 . Therefore, the value of c is 31 .

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