If the equation x2 - bx - 1 = 0 does notpassess real roots, then

1.	If the equation x2 - bx - 1 = 0 does notpassess real roots, then
Answer» a= 1, b= -b , C= -1condition for real root =b^2-4ac >0(-b)^2-4(1)(-1)>0b^2+4>0 David is correct. Following up, one realizes that b^2 = 24a at the solution. Thus, a = b^2/24 Then, 3a + b = 3 * b^2/24 + b = b^2/8 + b We minimize either by completing the square or taking the first derivative. Completing the square, b^2/8 + b = 1/8(b^2 + 8b) = 1/8(b^2+8b+16) - 1/8(16) = 1/8(b+4)^2 - 2 This is minimized at b = -4. a = b^2/24 = (-4)^2/24 = 2/3 As a check, note that 2/3 x^2 - 4 x + 6 = 2/3 (x^2 - 6x + 9) = 2/3 (x - 3)^2 has one solution, not two. a=1; b=-1; c=-1, b^2-4ac=(-1)^2-4(1)(-1); =1+4(1)=5

Answer»

a= 1, b= -b , C= -1condition for real root =b^2-4ac >0(-b)^2-4(1)(-1)>0b^2+4>0

David is correct. Following up, one realizes that b^2 = 24a at the solution. Thus, a = b^2/24

Then, 3a + b = 3 * b^2/24 + b = b^2/8 + b

We minimize either by completing the square or taking the first derivative.

Completing the square, b^2/8 + b = 1/8(b^2 + 8b) = 1/8(b^2+8b+16) - 1/8(16) = 1/8(b+4)^2 - 2

This is minimized at b = -4. a = b^2/24 = (-4)^2/24 = 2/3

As a check, note that 2/3 x^2 - 4 x + 6 = 2/3 (x^2 - 6x + 9) = 2/3 (x - 3)^2 has one solution, not two.

a=1; b=-1; c=-1, b^2-4ac=(-1)^2-4(1)(-1); =1+4(1)=5

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