Given that ax² + bx + c = 0 has no real roots and a + b + c

1. c = 0
2. c > 0
3. c
4. None of these

will u please complete the question

the data is not sufficient

 

1. A.c = 0
2. B.c > 0
3. C.c
4. D.None of these
5. This is the question :-)

Need the answer reallyfast. PLEASE.

what about the third option

just check out that the dicriminant should be less than 0 of the non real roots

b² −4ac

c less than 0 is the third option

 

I kind of deducted that b²-4ac ²² is always positive...so “ac” should also be positive, implying that a and c should bopth either be positive or negative. I don’t know if I am right or not

damn the keyboard. I mean b^2 – 4ac always less than 0 so since b^2 is always positive, ac has to be positive

of course

an example is here

 

-8x² +4x-1

i think that the example is fullfilling all the conditions of your question just check out

Can you somehow prove it graphically?

over here image cannot be posted be a student

sorry

can you prove it mathematically? Your exapmle fits perfectly but I would really like to know the concept behind the question.

ok let me tell u then

do

u know Descartes' rule of signs

search it out on google or ask your teacher than the problen is direct

nothing to be confusing just check it out

answer is option c

if a+b+c is less than 0 that means that f(1) is less than  zero 

so  since the given quartatic has no real roots  the whole graph (parabola) lies below x axis 

so f(x)

so c=f(0) is also less than zero

PLZ APPROVE

 

in your case don’t apply descarets law because this law is for determining the nature of roots of higher degree equations

in your case don’t apply descarets law because this law is for determining the nature of roots of higher degree equations

If there are no real zeros, the entire graph is either above the x-axis or below it. Given that the function value at x =1 is negative, the entire function is indeed below the x-axis. Therefore, the function value at x = 0, i.e., c is also below the x-axis and hence c is negative.