Deciding values for b?

Question

For 5x^2+bx+1=0
Deciding for all values of b, I came up with b=2 or -2. Does anyone have anything different?

Answer 1

I like 17.

Answer 2

There are two unknown variables and you have only one equation. You can get each one in terms of the other but cannot get the right answer untill you have another equation.

Always remember, the number of equations needed to solve the equation is equal to the number of unknown variables in it.

If you get the second equation, try solving it simultaneously and you should get the answer. By this I mean, Try to get an equation in terms of only x or only b by substituting from the second equation and then solve. I wish this will help you out.

Answer 3

when you factor this into (5x+1)(x+1) you should get 6 as you b value.

Answer 4

observe that those equations are all quadratics (the optimal exponent of x is two). therefore, they could all be solved through ability of the quadratic formula: for a quadratic ax² + bx + c, x = (-b ± sqrt(b² - 4ac))/2a So, reckoning on the quadratic formula we see the time period well-known through actuality the discriminant, b² - 4ac. on condition that's interior a sq. root, suitable that's declared. If: b² - 4ac = 0, there is a million genuine root (the sq. root is 0, and we are left with -b/2a interior the quadratic formula) b² - 4ac > 0, there are 2 genuine roots, through ±. b² - 4ac < 0, there aren't any further any genuine roots, on account that you've gotten the sq. root of a adverse sort, that's imaginary.

Answer 5

(5x+1)(x+1), 6
(5x-1)(x-1), -6
b = -(5x^2+1)/x
yields integer values of only 6 and -6

Answer 6

You cant really do it until you get another equation. Its 2 variables X & B

Answer 7

You have too many variables to determine b. Is there a second equation that goes with this one?

Answer 8

You don't understand the problem! Any value of b is ok there.

Answer 9

32 works for me.

Answer 10

nope, you're right