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

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

got it.. its 1,4

hold on

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

the equation should be \[(2(k+2))^2-36=0\]

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

which gives 1 and -5

thank u :)

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

\[4(k+2)^2=36\] \[(k+2)^2=9\] \[k+2=3\] \[k+2=-3\] \[k=1\] \[k=-5\]

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

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

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

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...

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

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