Equal Real Roots

**casey_k** · January 24th 2009, 07:40 AM

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.

**Mush** · January 24th 2009, 08:06 AM

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.

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

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

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

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

The discriminant must be zero for equal roots:

$b^2-4ac = 0$

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

$b = (-k-5)$

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

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

**Danny** · January 24th 2009, 08:29 AM

Originally Posted by Mush

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

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

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

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

The discriminant must be zero for equal roots:

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

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

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

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

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

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

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