SOLUTION: hey,
could you please solve the following question, thanks,
The graph of function y= x^2 - kx + k + 8 touches the x-axis at one point. What
is the value of k?
I will be wait
Algebra.Com
Question 160346: hey,
could you please solve the following question, thanks,
The graph of function y= x^2 - kx + k + 8 touches the x-axis at one point. What
is the value of k?
I will be waiting for the solution,
and thanks again,
Found 2 solutions by stanbon, ankor@dixie-net.com:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
The graph of function y= x^2 - kx + k + 8 touches the x-axis at one point. What
is the value of k?
------------------------------
A quadratic has one solution when the discrimanant is zero.
-----------------
Your quadratic has a = 1, b = -k,c = k+8
---------
Discrimant equation:
b^2 - 4ac = 0
(-k)^2-4*1*(k+8) = 0
k^2 -4k -32 = 0
k^2-8k+4k-32 = 0
k(k-8)+4(k-8) = 0
(k-8)(k+4) = 0
k = 8 or k = -4
==============================
Cheers,
Stan H.
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The graph of function y= x^2 - kx + k + 8 touches the x-axis at one point. What
is the value of k?
:
we use the discriminant: b^2 - 4*a*c = 0; when it touches one point on the x axis (a double root)
:
in this equation a=1, b=k, c=(k+8)
Substitute:
k^2 - 4*1*(k+8) = 0
k^2 - 4k - 32 = 0
Factor
(k-8)(k+4) = 0
Two solutions
k = 8
k = -4
:
for k=8
x^2 - 8x + 8 + 8 = 0
x^2 - 8x + 16 = 0; which is(x-4)^2 a double root at x=4
and for k=-4
x^2 -(-4)x + (-4) + 8 = 0
x^2 + 4x + 4 = 0; which is(x+2)^2 a double root at x=-2
:
:
Did that help?
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