4x^2-29x+7=0??
do you know the quadratic formula ?
no, I usually complete the square.
I need to solve it using completing the square.
So first you wld multiply a*c which is 4*7=28 Then we need to find two numbers that when you add them equal b=-29 and when you multiply it it equals A*C=28 So the 2 numbers wld be -1 and -28 -1*-28=28 -1-28=-29
ohhh sorry i didnt realize that
wait nearly finished solving it give me one more sec
kkk i figured it out but it is sooooo ugly
f(x)=4(x-(29/8))^2-(729/16)
okkk ill show u the steps now if u wld like
no I am here
ok so the first step is to kind of separate the number with the x^2 and the x leaving the 7 alone and out of the bracket f(x)=4(x^2-(29/4)x)+7
next we take our middle termwhich -(29/4) and we divide it by 2 and then square it (29/4)/2=(29/8) (29/8)^2=841/64 f(x)=4(x^2-(29/4)x+(841/64))+7 Now we know that whtvr we do to one side you have to do to the other. So since we added 4(841/64) to one side we have to add it to the other side f(x)+4(841/64)=4(x^2-(29/4)x+(841/64))+7 f(x)=4(x^2-(29/4)x+(841/64))+7-(4*(841/64)) f(x)=4(x^2-(29/4)x+(841/64))-(729/16)
ok the last step is to take the middle bracket and "complete the square" Wld u by any chance know how to do that?
ok well i tell u what to do in this case so u will know what to do next time you take the midlle term that u divided by two which was (29/4)/2=29/8 that term when squared = (841/64) and when added twice equals (29/4) Remember we leave the sign that it had originally which was a negative f(x)=4(x-(29/8))^2-(729/16) I may have confused u along the way but in general u dont have to understand the logic you just have to follow a set of rules. wait let me see if i can find u a link to the set of steps that u have to follow
http://www.mathsisfun.com/algebra/completing-square.html ok I like this one and they solved it in an easier way than i did lol scroll down to the bottom and like ull see this highlighted peach box and like there they give u the steps like its very easy too follow
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