Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
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Postby cobbler1516 » Sun Dec 23, 2012 8:47 pm

Can anyone help me - find an answer for b? I need a real number answer. Thanks

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Re: x^2-bx+18=0

Postby maggiemagnet » Sun Dec 23, 2012 11:36 pm

The math hasn't changed sinc your other post: with two variables and only one equation it is not possible to get a number answer for this. Sorry! :oops:

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Re: x^2-bx+18=0

Postby jg.allinsymbols » Sat Dec 29, 2012 3:18 am

Solving for just b, do these first:

*Add +bx to both sides,
*Multiply both sides by (1/x)

That will give you b as a function of x, which seems to be a variable, possibly of two complex numbers. If these values for x are to be in the form, a1+a2i, then you want a2=0, since you specified wanting Real number solutions.

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Re: x^2-bx+18=0

Postby alan » Fri Jan 04, 2013 12:28 am

The above post is correct, add bx to both sides and multiply by 1/x to get


Plug in any value of x to get a valid value for b. For example when x=1 then b=19 so x^2 -19x+18=0 is a solution. This solution is factorable as (x-1)(x-18).
Another solution is x=-1 which gives b=-19 and the quadratic is x^2+19x+18=0. In factored form this is (x+1)(x+18)
x= + or - 2 also gives the factorable quadratics x^2-11x+18=0 and x^2+11x+18=0 which factor as (x-2)(x-9) and (x+2)(x+9).

Pick any value for x and the two factors for the factored form of the quadratic will be -x and -18/x. The two numbers always multiply to give 18 and they add to give b.

Another example is x=1/2, then 18/x=36, the factored form of the quadratic is (x-1/2)(x-36)=0. In this case b=1/2+18/(1/2)=1/2+36=73/2
which gives x^2-(73/2)x +18 which factors to (x-1/2)(x-36). Notice that the factors are -x and -x/18 as expected.

So we can now see that the answer to the original question of finding the solution to x^2-bx+18=0 is x=n and x=18/n for any value of n where n is a rational number.

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