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@hartnn
for the equation \(\Large ax^2+bx+c = 0\) to have no real roots, \(\large b^2 -4ac\) must be less than 0. so first compare x^2 +2x+c = 0 with ax^2 +bx+c = 0 and find a,b,c
a = 1, b = 2 c =1
a= 1, b = 2 and c = c now find b^2 -4ac
2^2 - 4(1)(1) = 0
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oh wait. c = c?
yes, keep c as c only
then it equals -4c+4?
\(\Large b^2-4ac < 0 \\ \large 4 - 4c < 0\) factor out the 4
or just add 4c on both sides
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4 < 4c divide both sides by 4
4(1-c)?
yea, ok 4 (1-c) < 0 dividing both sides by 4 1-c < 0 or 1<c
which option satisfies that c>1 ??
there will be no real roots when c is greater than 1
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so 0?
is 0 greater than 1 ?
3.
good! \(\huge \checkmark \)
ty
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welcome ^_^
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