SOLUTION: For what values of k will the equation {{{ x^2 -(k-2)x-12=0}}} have roots that are equal but opposite in sign

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Question 1013006: For what values of k will the equation +x%5E2+-%28k-2%29x-12=0 have roots that are equal but opposite in sign
Found 3 solutions by ikleyn, Theo, MathLover1:
Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.
For k = 2.

In this case the line x = 0 is the symmetry line for the parabola.

You can deduce it from the quadratic formula too.


Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
when k = 2, the equation of x^2 - (k-2)x - 12 = 0 becomes x^2 - -0x - 12 = 0 which becomes x^2 - 12 = 0.
add 12 to both sides of this equation to get x^2 = 12.
take the square root of this equation to get x = plus or minus sqrt(12).





Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

+x%5E2+-%28k-2%29x-12=0
to have roots that are equal but opposite in sign, we can write our equation like this:
+x%5E2+-%28k-2%29x-12=0
+%28-x%29%5E2+-%28k-2%29%28-x%29-12=0
And since we started with +x%5E2+-%28k-2%29x-12=0
we can set the left sides equal:
+x%5E2+-%28k-2%29x-12=++%28-x%29%5E2+-%28k-2%29%28-x%29-12+....solve for k
+x%5E2+-kx%2B2x-12=+x%5E2+%2Bkx-2x-12+

+-k%2B2=+k-2+
+2%2B2=+k%2Bk+
+4=+2k+
k=2

check:
+x%5E2+-%28k-2%29x-12=0
+x%5E2+-%282-2%29x-12=0
+x%5E2+-%280%29x-12=0
+x%5E2-12=0
+x%5E2=12
+x=sqrt%2812%29
+x=sqrt%284%2A3%29
+x=2sqrt%283%29
solutions: +x=2sqrt%283%29 and +x=-2sqrt%283%29