# Equal roots

• Jun 13th 2008, 11:01 AM
storchfire1X
Equal roots
Hi! I need some help on understanding this proof:

If the equation $x^2 + 2(k+2)x +9k = 0$ has equal roots, find k.

The condition for equal roots gives
$(k+2)^2 = 9k$
$k^2 - 5k + 4 =0$
$(k-4)(k-1) = 0$
$k=4 , k=1$

The proof makes sense: When k = 4, for example, we get $x^2 + 12x + 36$ and this will give equal roots if $b^2 - 4ac = 0$ which it does since $144 - 144 = 0$

But I still don't understand where the first line of the proof,
$(k+2)^2 = 9k$, comes from

From Higher Algebra, H.S Hall & Knight
• Jun 13th 2008, 11:05 AM
TKHunny
Do you remember "Completing the Square"? It is just a step from that process, usually verbalized, "The square of one-half the coefficient of the linear term". Probably, it is why the '2' is part of the linear term, so that finding half of it will be more convenient.
• Jun 13th 2008, 11:19 AM
o_O
Here: $a = 1$, $b = 2(k+2)$, $c = 9k$.

Plug this into the discriminant:
$b^{2} - 4ac = 0$
$\big[2(k+2)\big]^2 - 4(1)(9k) = 0$
$4(k+2)^{2} = 36k$ (squared the first term recalling that $(ab)^{n} = a^nb^n$, moved the 2nd term to the other side)
$(k+2)^2 = 9k$
• Jun 13th 2008, 11:40 AM
Reckoner
Quote:

Originally Posted by storchfire1X
The proof makes sense: When k = 4, for example, we get $x^2 + 12x + 36$ and this will give equal roots if $b^2 - 4ac = 0$ which it does since $144 - 144 = 0$

You said it yourself: the equation will have equal roots iff the discriminant $b^2 - 4ac = 0$.

So, we have

$b^2 - 4ac = 0$

$\Rightarrow\left[2\left(k + 2\right)\right]^2 - 4(1)(9k) = 0$

$\Rightarrow4\left(k + 2\right)^2 = 4(9k)$

$\Rightarrow\left(k + 2\right)^2 = 9k$

Edit: Beaten by 21 minutes! Ouch.