# Equal Real Roots

• Jan 24th 2009, 07:40 AM
casey_k
Equal Real Roots
f(x) = -1/2x^2 -5x +4

find the value(s) of k such that the equation f(x+k)=0 has two equal real roots.
• Jan 24th 2009, 08:06 AM
Mush
Quote:

Originally Posted by casey_k
f(x) = -1/2x^2 -5x +4

find the value(s) of k such that the equation f(x+k)=0 has two equal real roots.

$\displaystyle f(x+k) = \frac{-1}{2}(x+k)^2 -5(x+k) + 4$

$\displaystyle f(x+k) = \frac{-1}{2}(x^2+2kx+k^2) -5x-5k + 4$

$\displaystyle f(x+k) = \frac{-1}{2}x^2-\frac{1}{2}2kx-\frac{1}{2}k^2 -5x-5k + 4$

$\displaystyle f(x+k) = \frac{-1}{2}x^2+x(-k-5) - \frac{k^2}{2} -5k + 4$

The discriminant must be zero for equal roots:

$\displaystyle b^2-4ac = 0$

$\displaystyle a = \frac{-1}{2}$

$\displaystyle b = (-k-5)$

$\displaystyle c = - \frac{k^2}{2} -5k + 4$

Insert these into the equation. Expand. Then solve for k.
• Jan 24th 2009, 08:29 AM
Jester
Quote:

Originally Posted by Mush
$\displaystyle f(x+k) = \frac{-1}{2}(x+k)^2 -5(x+k) + 4$

$\displaystyle f(x+k) = \frac{-1}{2}(x^2+2kx+k^2) -5x-5k + 4$

$\displaystyle f(x+k) = \frac{-1}{2}x^2-\frac{1}{2}2kx-\frac{1}{2}k^2 -5x-5k + 4$

$\displaystyle f(x+k) = \frac{-1}{2}x^2+x(-k-5) - \frac{k^2}{2} -5k + 4$

The discriminant must be zero for equal roots:

** $\displaystyle b^2-4ac = 0$

** $\displaystyle a = \frac{-1}{2}$

** $\displaystyle b = (-k-5)$

** $\displaystyle c = - \frac{k^2}{2} -5k + 4$

Insert these into the equation. Expand. Then solve for k.

Problem is that no $\displaystyle k$ will solve this set of equations (**). If you look at the graph of $\displaystyle f(x)$, $\displaystyle f(x+k)$ only translates the graph left and right. There will always be two real roots (never repeated). Now if the question read $\displaystyle f(x)+k$ then we could translate up and down to get a repeated real root.