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How do you find the value of K so each given trinomial is a perfect square? x^2+12x+k
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36
how do you get that? :O
12/2 = 6 6^2 = 36
could i do the same with this 9x^2-10x+k ?
you can do it but you will not get an integer
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you will get a fraction if you want a perfect square
the way you can make sure you have a perfect square is to make sure that \[b^2-4ac=0\] so in this case you want \[(-10)^2-4\times 9\times k=0\] or \[100=36k\] \[k=\frac{100}{36}=\frac{25}{9}\]
and then you will get \[9x^2-10x+\frac{25}{9}=\frac{1}{9}(9x-5)^2\]
Oh I see, Thank you! :)
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