Question

if the roots of the equation (b-c)x^2+(c-a)x+(a-b)=0 are equal then prove that 2b=a+c

Answer 1

Given :

• roots of equation (b-c) x² + (c-a) x + (a-b) = 0 are equal

To prove :

• 2 b = a + c

Knowledge required :

A quadratic equation in its standard form ,i.e, a x² + b x + c = 0 will have equal and real roots when,

$\red{\star}\;\boxed{\sf{Discriminant=b^2-4ac=0}}$

Solution :

As given (b-c) x² + (c-a) x + (a-b) = 0 have equal roots therefore,

$\red{\implies}\sf{(c-a)^2 - 4 (b-c)(a-b) = 0}$

$\red{\implies}\sf{(c^2 + a^2 - 2 ac - 4 (ab - ac - b^2 + bc ) = 0}$

$\red{\implies}\sf{c^2 + a^2 - 2 ac - 4ab +4 ac +4 b^2 -4 bc = 0}$

$\red{\implies}\sf{ a^2 + 4 b^2 + c^2 + 2 ac - 4ab -4 bc = 0}$

$\red{\implies}\sf{ ( a)^2 + (-2 b)^2 + (c)^2 + 2 (a)(c) +2(a)(-2b) +2(-2b)(c) = 0}$

$\red{\star \sf{using\:algebraic\:identity}}$

$\boxed{\red{\sf{(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz}}}$

$\red{\implies}\sf{(a-2b+c)^2=0}$

$\red{\implies}\sf{a-2b+c=0}$

$\red{\implies\sf{a+c=2b}}$

Proved .

Answer 2

Answer:

➡Given :

✏Roots of equation

= (b-c) x² + (c-a) x + (a-b) = 0 are equal

➡To prove :

✏2 b = a + c

➡Knowledge required :

A quadratic equation in its standard form ,i.e,

a x² + b x + c = 0 will have equal and real roots when,

⭐ Discriminant = b² - 4ac = 0⭐

➡Solution :

As given (b-c) x² + (c-a) x + (a-b) = 0 have equal roots therefore,

=> (c-a)² - 4(b-c)(a-b) = 0

=> (c²+a²-2ac-4(ab-ac-b²+bc) =0

=> c²+a²-2ac-4ab+4ac+4b²-4bc = 0

=> a²+4b²+c²+2ac-4ab-4bc = 0

=> (a)²+(-2b)²+(c)²+2(a)(c)+2(a)(-2b)+2(-2b)+2(-2b)(c) = 0

⭐ Using algebraic identity:-

➡(x+y+z)² = x²+y²+z²+2xy+2yz+2xz

=> (a-2b+c)² = 0

=> a-2b+c = 0

=> a+c = 2b

⭐ Proved ⭐

Step-by-step explanation:

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