Math, asked by legendsayan761, 6 months ago

(b-c)x²+(c-a)x+(a-b)=0 সমীকরণে বীজদ্বয় সমান হলে দেখাও যে 2b=a+c​

Answers

Answered by greeshu2005
3

Answer:

(b-c)x²+(c-a)x+(a-b)=0

Comparing with quadratic equation

Ax²+Bx+C=0

A=(b-c),B=c-a,C=a-b

Discriminate when roots are equal

D=B²-4AC=0

D=(c-a)²−4(b-c)(a-b)=0

D=(c²+a²−2ac)-4(ba-ac-b²+bc)=0

D=c²+a²−2ac-4ab+4ac+4b²-4bc=0

c²+a²+2ac-4b(a+c)+4b²=0

(a+c)²-4b(a+c)+4b²=0

[(a+c)-2b]²=0

a+c=2b

If the roots of the equation (b-c) x^2 +(c-a) x+(a-b) =0 are equal, the prove that 2b=a+c.(?)

What are the roots of the equation (b-c)x^2+(c-a)x+(a-b)=0?

For any vectors A, B and C, prove that (A*B).(B*C)*(C*A)=/(A. B)*C)/2. How can I prove it?

If x + b > c and x > 0 have the same solutions, what is the relationship between b and c?

If the quadratic equation (b-c)x^2+(c-a) x+(a-b) =0 has equal roots, what is a+c=?

A2a!!

If roots of a quadratic equation are equal, then the discriminant of the quadratic equation is 0.

D=b2−4ac=0

(b−c)x2+(c−a)x+(a−b)=0 ≡ ax2+bx+c=0

Here, a =(b-c) , b = (c-a) and c = (a-b)

So, D = (c−a)2−4(b−c)(a−b)=0

c2+a2−2ac−4(ab−b2−ac+bc)=0

c2+a2−2ac−4ab+4b2+4ac−4bc=0

c2+a2+2ac+4b2−4ab−4bc=0

(c+a)2+4b2−4b(a+c)=0

(c+a)2+(2b)2−2⋅(c+a)⋅(2b)=0

[(c+a)−(2b)]2=0

⇒ c+a−2b=0

a + c = 2b

Hence proved...

Step-by-step explanation:

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