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Mathematics 11 Online
OpenStudy (anonymous):

how do i get that answer? 0_o

OpenStudy (anonymous):

OpenStudy (anonymous):

Well, (-1/2)/2 is the same as (-1/2) * (1/2)

OpenStudy (anonymous):

Now square that.

hartnn (hartnn):

division by 2 means multiplication by 1/2 so (-1/2)*(1/2)=-1/4 now square it.

OpenStudy (anonymous):

1/16... so -31/2 it would be (-31/2)*(31/2)=-961/4?

OpenStudy (anonymous):

I'm not sure what the second part is there at all, but (-31/2)^2 = 240.25 in decimal form.

OpenStudy (anonymous):

Is that some separate equation? part of another problem? what is that?

OpenStudy (anonymous):

yes let me take a pic

OpenStudy (anonymous):

oh wait

hartnn (hartnn):

keep it as 961/4 only.

OpenStudy (anonymous):

hartnn (hartnn):

(1/7) should be multiplied to all 3 terms and not only to 7x^2

OpenStudy (anonymous):

ooh and do the same thing with -217? right?

hartnn (hartnn):

where did u get -217 from ?

OpenStudy (anonymous):

wait do i multiply by 7 or just -37/7?

hartnn (hartnn):

why is there need to multiply 1/7 on both sides ?

OpenStudy (anonymous):

so i can get rid of the 7 and be left with x^2

OpenStudy (ash2326):

@cos00155079 you want to create a perfect square term \[7x^2-31x-20=0\] multiply both sides by 1/7 or divide by 7 \[\frac{1}{7}(7x^2-31x-20)=0\times \frac 17\] we get \[x^2-\frac {31}7 \ \ \ x-\frac {20}{7}\ =0\] do you get this?

OpenStudy (anonymous):

yup

OpenStudy (ash2326):

now half the x term and add and subtract the square of it \[\frac{-31}{7}\\ \div 2=\frac{-31}{14}\] square it \[{(\frac{-31}{14})}^2\] now add and subtract this on the left side, do you get this?

OpenStudy (anonymous):

yes!! let me solve it from here..try to..

OpenStudy (ash2326):

good, take your time:) let me know if you need help :)

OpenStudy (anonymous):

ok but if i square that -31/14 that gives me 961/196???

OpenStudy (ash2326):

yeah, you're right

OpenStudy (anonymous):

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