if roots of equation (b-c)x²+(c-a)x+(a-b)=0 are equal,then show that 2b=a+c
Answers
Step-by-step explanation:
Given :-
(b-c)x^2+(c-a)x+(a-b)=0 has equal roots
To find :-
Show that 2b = a + c .
Solution:-
Given quardratic equation is
(b-c)x^2+(c-a)x+(a-b)=0
On Comparing this with the standard quadratic equation ax^2+bx+c = 0
We have
a = (b-c)
b = (c-a)
c = (a-b)
Given that
It has equal roots.
=> It's discriminant must be zero
We know that
The discriminant of a quardratic equation ax^2+bx+c = 0 is D=b^2-4ac
Now,
=> b^2-4ac = 0
=> (c-a)^2 -4(b-c)(a-b) = 0
=> (c^2-2ac+a^2) -4(ab-b^2-ac+bc) = 0
Since (a-b)^2 = a^2-2ab+b^2
=> c^2-2ac+a^2-4ab+4b^2+4ac-4bc = 0
=> c^2+a^2+4b^2+(-2ac+4ac)-4ab-4bc = 0
=> c^2+a^2+4b^2+2ac-4ab-4bc = 0
It can be rearranged as
=> c^2+a^2+4b^2-4ab-4bc+2ac = 0
It can be written as
=> (a)^2+(-2b)^2+(c)^2+2(a)(-2b)+2(-2b)(c)+2(c)(a)=0
It is in the form of x^2+y^2+z^2+2xy+2yz+2zx
Where
x = a
y = -2b
z = c
We know that
(x+y+z)^2 =x^2+y^2+z^2+2xy+2yz+2zx
=>(a)^2+(-2b)^2+(c)^2+2(a)(-2b)+2(-2b)(c)+2(c)(a)=0
=> [a+(-2b)+c]^2 = 0
=> a+(-2b)+c = 0
=> a+c -2b = 0
=> a+c = 2b
=> 2b = a+c
Hence, Proved.
Used formulae:-
- The standard quadratic equation is ax^2+bx+c = 0
- The discriminant of a quardratic equation ax^2+bx+c = 0 is D=b^2-4ac
- (a-b)^2 = a^2-2ab+b^2
- (x+y+z)^2 =x^2+y^2+z^2+2xy+2yz+2zx
More to know:-
- If D= b^2-4ac≥0 then the real roots exists.
- If D> 0 , it has two distinct and real roots.
- If D<0 it has no real roots i.e.imaginary roots.
- If D=0 it has two equal and real roots.
Answer:
hello
if you don't mind
can I be your friend
if yes then
Step-by-step explanation:
your introduction