Help in completing the square!
\[\large 2v^2 -20v - 7 = 7\]
@TheSmartOne
\[v=5\pm4\sqrt{2}\]
\(\sf\large a^2 -2ab + b^2 = (a-b)^2\) So we have a = 2 and 2ab = 20 so solve for b 2(2)b = 20 b = >
b = ? *
\[2v^2 - 20v - 14 = 0\] divided everything by 2 \(v^2 - 10v - 7 = 0\) \(v^2 - 10v + (-10/2) = 7 + (-10/2)\) \(v^2 - 10v - 5 = 7 - 5\) \(\sqrt{v^2 - 10v - 5} = \sqrt{2}\) I know that i will be getting a very weird answer from the answer key.
is there anything that i am doing wrong?
mhmm
it should be (-10/2)^2
because we have to add b^2 on both sides so we could simplify it :)
but for a previous question, i used the same thing that i am doing now and it matched with the answer key.
but this time it isn't matching the answer key? o_0
Co incidence ,or you match it wrong, it should be square of -5
okay i understand what you're saying. for the last one it was \((2/2)^2 = 1^2 = 1\) that's why i didn't understand that :)
great :)
thanks!
wait a second
ok
nevermind, you got it correct :)
hold on a moment
so i did: \(v^2 - 10v + 25 = 7 + 25\) \(v^2 - 10v + 25 = 32\) \(32\) is not a perfect square but the answer key has whole numbers
@TheSmartOne
so solving that we get \(\sf\Large v - 5 = \pm\sqrt{32}\) \(\sf \Large v = 5 \pm \sqrt{32}\) \(\sf\Large v = 5\pm 4\sqrt{2}\) what does the answer key say the answer is?
2v^2 - 12v - 7 = 7 {7,-1} <--- answer
do you know my mistake?
maybe we shouldn't have simplified
not sure really mhmm :/
what about this problem \(6v^2 - 12v - 84 = 6\)
WAIT
\(\color{blue}{\text{Originally Posted by}}\) @calculusxy 2v^2 - 12v - 7 = 7 {7,-1} <--- answer \(\color{blue}{\text{End of Quote}}\) we have -12v here not -20v as you posted
oh wow!
:)
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