Thread: find the 2nd root

X^2-11x+k=0 one of the roots=5 find the other root.

Hello,

Originally Posted by blame_canada100

X^2-11x+k=0 one of the roots=5 find the other root.

If a and b are the roots of a polynomial x²+mx+n=0, then x²+mx+n=(x-a)(x-b)
If you develop the RHS, you'll get x²-(a+b)x+ab.
So you can see that m=-(a+b)

Applying it here, you'll get -11=-(5+other root)

Hi,

One place to start is to use the basic quadratic formula:

x=(-b+/-(b^2-4ac)^0.5)/2a.

With a=1, b=-10, and c=k, and letting x=5 (as one of the roots) gives you k=25.

Substituting this back into the question gives you x^2-10x+25=0. This is easily factorised to give x=5, twice.

So the answer to the question is that the other solution is also 5.

you can use discriminant X(squared)-11x+k=aX(squared)+bx+c

a=1, b=-11, c=k, and let's say the discriminant D

D=b(squared)-4ac
=121-4*1*k
=121-4k

solutions are s1 and s2

S1=(-b+(square root)D)/2a

=(-11+(square root)121-4k)/2

and S1=5...so
(-11+(square root)121-4k)/2=5
-11+(square root)121-4k=10

(square root)121-4k=21
now you put everywhere by squared and you get
121-4k=441
-4k=320
k=-80

so now let's look for S2
D=(square root)121-4(-80)
D=(square root)121+320
D=(square root)441
D=21

S2=(-b-D)/2a
S2=(-11-21)/2
S2=-16
the second root is -16