If the roots of the quadratic equation
(b - c)x2 + (c - a)x + (a - b) = 0 are equal, prove 2b = a + C.
Answers
ANSWER
If roots of a quadratic equation are equal, then discriminant of the quadratic equation is 0
D=b
2
−4ac=0
(b−c)x
2
+(c−a)x+(a−b)=0
Comparing with
ax
2
+bx+c=0
Here, a=(b−c), b=(c−a) and c=(a−b)
So,
⇒(c−a)
2
−4(b−c)(a−b)=0
⇒c
2
+a
2
−2ac−4(ab−b
2
−ac+bc)=0
⇒c
2
+a
2
−2ac−4ab+4b
2
+4ac−4bc=0
⇒c
2
+a
2
+2ac+4b
2
−4ab−4bc=0
⇒(c+a)
2
+4b
2
−4b(a+c)=0
⇒(c+a)
2
+(2b)
2
−2(c+a)(2b)=0
⇒[(c+a)−(2b)]
2
=0
⇒c+a−2b=0
⇒2b=c+a
Answer:
Step-by-step explanation:
ANSWER
If roots of a quadratic equation are equal, then discriminant of the quadratic equation is 0
D=b
2
−4ac=0
(b−c)x
2
+(c−a)x+(a−b)=0
Comparing with
ax
2
+bx+c=0
Here, a=(b−c), b=(c−a) and c=(a−b)
So,
⇒(c−a)
2
−4(b−c)(a−b)=0
⇒c
2
+a
2
−2ac−4(ab−b
2
−ac+bc)=0
⇒c
2
+a
2
−2ac−4ab+4b
2
+4ac−4bc=0
⇒c
2
+a
2
+2ac+4b
2
−4ab−4bc=0
⇒(c+a)
2
+4b
2
−4b(a+c)=0
⇒(c+a)
2
+(2b)
2
−2(c+a)(2b)=0
⇒[(c+a)−(2b)]
2
=0
⇒c+a−2b=0
⇒2b=c+a