If the equation x2 + 2(k+2)x + 9k = 0 has equal roots, find k?

Question

anonymous · Answer

set 
$$(2k+2)^2-4\times 9=0$$ solve for k

anonymous · Answer

got it.. its 1,4

anonymous · Answer

hold on

anonymous · Answer

i wrote it wrong but when you square don't forget to square the 2

anonymous · Answer

the equation should be 
$$(2(k+2))^2-36=0$$

anonymous · Answer

i got t by jus bruteforce approach.. jus substituted de val fr k and found t

anonymous · Answer

which gives 1 and -5

anonymous · Answer

thank u :)

anonymous · Answer

The motivation is that you want your quadratic to be of the form$$x^{2}+2bx+b^{2} = (x+b)^{2}$$
That means that (k+2)=b and 9k=b^2
So, (k+2)^2=9k...solve for k

anonymous · Answer

$$4(k+2)^2=36$$
$$(k+2)^2=9$$
$$k+2=3$$
$$k+2=-3$$
$$k=1$$
$$k=-5$$

anonymous · Answer

So $$k^{2}+4k+4=9k$$
$$k^{2}-5k+4=0$$
$$(k-4)(k-1)=0$$
k=1, 4

anonymous · Answer

@mtbender nicely put! i was thinking routine discriminant = 0 but that is a better view

anonymous · Answer

the condition for an equation to have equal roots is,
b^2-4*a*c=0;
applying this condition to the equation,taking,
a=1,b=2*(k+2),c=9*k
we get k value as,1 and 4

anonymous · Answer

Ahh...ok @satellite.  Looking at @sravan's post, you missed the k on the right side of the equals.  That's why you got -5 instead of 4.  I was trying to figure out why we had different answers...

anonymous · Answer

oh heavens yoy are right. i wrote -36 should be -36k!