1. ## 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

2. 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.

3. 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)
$(k+2)^2 = 9k$

4. 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.