If the equation x squared minus bx + q = 0 and x squared minus x + b = 0 have a common road and the second equation has equal roots then?

1 Answer
Jun 14, 2018

b=1/4, q=-1/8

Explanation:

The question isn't clearly phrased. I think what's being asked is:

Take the two-equation system x^2-bx+q=0 and x^2-x+b=0. If we know that the two equations have a common root, and that the second equation has two identical roots, then deduce b and q.

If the second equation has two equal roots, then it's of the form (x-alpha)^2=0, where the root is alpha. This multiplies out as
x^2-2alphax+alpha^2=0. Comparing this to the above second equation, we see that 2alpha=1 from the first coefficient (so alpha=1/2), and that alpha^2=b from the second coefficient (so b=1/4). Thus the second equation is x^2-x+1/4=0, or, factorised, (x-1/2)^2=0.

If the two equations have a common root, then one of the above roots must be a root of the first equation. But the two above roots are both equal, so we know the shared root is also x=1/2. If the other root is beta, then the equation has the form (x-1/2)(x-beta)=0, which multiplies out as x^2-(beta+1/2)x+1/2beta=0.
The x coefficient lets us deduce beta via beta+1/2=b=1/4. So beta=-1/4. Then we can deduce q from the constant coefficient: q=1/2beta=1/2*(-1/4)=-1/8. Thus the first equation is x^2-1/4x-1/8=0, or, factorised, (x-1/2)(x+1/4)=0.