Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

Given that ax 2 + bx + c = 0 has no real roots and a + b + c c = 0 c > 0 c None of these

Given that ax+ bx + c = 0 has no real roots and a + b + c
  1. c = 0
  2. c > 0
  3. c
  4. None of these

Grade:12th Pass

20 Answers

grenade
2061 Points
6 years ago
will u please complete the question
the data is not sufficient
Anjali Singh
23 Points
6 years ago
 
Given that ax+ bx + c = 0 has no real roots and a + b + c
  1. A.c = 0
  2. B.c > 0
  3. C.c
  4. D.None of these
  5. This is the question :-)
Anjali Singh
23 Points
6 years ago
Given that ax+ bx + c = 0 has no real roots and a + b + c less than 0
Anjali Singh
23 Points
6 years ago
Need the answer reallyfast. PLEASE.
grenade
2061 Points
6 years ago
what about the third option
grenade
2061 Points
6 years ago
just check out that the dicriminant should be less than 0 of the non real roots
grenade
2061 Points
6 years ago
b2 −4ac
Anjali Singh
23 Points
6 years ago
c less than 0 is the third option
 
I kind of deducted that b2-4ac 2is 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
Anjali Singh
23 Points
6 years ago
damn the keyboard. I mean b^2 – 4ac always less than 0 so since b^2 is always positive, ac has to be positive
grenade
2061 Points
6 years ago
of course
grenade
2061 Points
6 years ago
an example is here
 
-8x2 +4x-1
i think that the example is fullfilling all the conditions of your question just check out
Anjali Singh
23 Points
6 years ago
Can you somehow prove it graphically?
grenade
2061 Points
6 years ago
over here image cannot be posted be a student
sorry
Anjali Singh
23 Points
6 years ago
can you prove it mathematically? Your exapmle fits perfectly but I would really like to know the concept behind the question.
grenade
2061 Points
6 years ago
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
Nicho priyatham
625 Points
6 years ago
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
Nicho priyatham
625 Points
6 years ago
PLZ APPROVE
 
grenade
2061 Points
6 years ago
in your case don’t apply descarets law because this law is for determining the nature of roots of higher degree equations
grenade
2061 Points
6 years ago
in your case don’t apply descarets law because this law is for determining the nature of roots of higher degree equations
Arnav Adhiya
15 Points
2 years ago
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.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free