solve for x : ax^2 + bx + c =0 when a + b + c = 0 & a,b,c are constants.?

Answer 1

a cannot be zero then itr is not quadratic

put b = -(a+c)
ax^2-(a+c) x + c =0
=>ax^2-ax-cx+c =0
=>ax(x-1) -c(x-1) =0
=>(ax-c)(x-1) =0
=> ax-c = 0 or x-1 =0
=> x = c/a or 1

Answer 2

X=C/A OR 1

Answer 3

We can write
[1] ... b = -(a + c)
and rewrite the first equation as
[2] ... ax^2 - (a+c)x + c = 0
This can be factored as
[3] ... (ax - c)(x - 1) = 0
So the solutions are
[4] ... x = c/a OR x = 1

Answer 4

For all real numbers a, b, and c:

If c = -(a + b), then x = 1 or x = -(a + b)/a, a ≠ 0

If b = -(a + c), then x = 1 or x = -c/a, a ≠ 0

If a = -(b + c), then x = 1 or x = -c/(b + c), b + c ≠ 0 or b ≠ -c

Answer 5

You have two equations and more than two unknowns. There are infinitely many solutions.

One solution is to simply have a, b, and c all equal zero.

If your a, b, and c are specified, then what you're asking for is simply the quadratic formula:
x = (-b plus minus sqrt(b^2-4ac))/2a.

Answer 6

stop giving this tenth class NCERT stuff. come on kiddo! I know u asked thios question just for funsake, ya ya i know u are knowing the Shreedharacharya's formula or the Quadratic formula.

Answer 7

there can be any no. of solutions coz values of a b and c can be anyting as long as they add upto 0

Answer 8

c=-a-b
ax^2+bx-a-b=0
a(x+1)(x-1)+b(x-1)=0
(x-1)(ax-a+bx-b)=0
x=1
or
x(a+b)=a+b
x=1
the solution is x=1,1.

Answer 9

X = (-(b)+/- Sqrt of (b²-4ac) )/ 2a

The law of decreminant

Answer 10

soln is given by:
x= (-b/2a)+- (b^2-4ac)^.5 /2a
if a+b+c=0, b=-(a+c),
i.e: x= (a+c)/2a +-(a-c)/2a

==> x=1, c/a
Hence soln: x=1,c/a