Question

If the roots of the quadratic equation
(a - b)x² + (b-c)x + (c - a) = 0 are equal, ther
prove that 2a = b + c.​

Answer 1

Given:-

Roots of the quadratic equation (a - b)x² + (b - c)x + (c - a) = 0 and are equal.

To Prove:-

2a = b + c

$\rule{100}2$

Proof:-

(a - b)x² + (b - c)x + (c - a) = 0

The above equation is in the form ax² + bx + c = 0

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

Given that roots are equal. Means it's determinant is also equal to zero i.e. D = 0

We know that

D = (b)² - 4ac = 0

⇒ (b - c)² - 4(a - b)(c - a) = 0

⇒ b² + c² - 2bc - 4[a(c - a) -b(c - a)] = 0

⇒ b² + c² - 2bc - 4(ac - a² - bc + ab) = 0

Take 2 common from (-4ac - 4a² - 4bc + 4ab)

⇒ b² + c² - 2bc - 4ac - 4a² - 4bc + 4ab = 0

⇒ b² + c² - 2bc - 2(2ac) + (2a)² + 4bc - 2(2ab) = 0

[ 4bc - 2bc = 2bc ]

⇒ b² + c² + 2bc - 2(2ac) + (2a)² - 2(2ab) = 0

⇒ b² + c² +/2bc - 2c(2a) + (2a)² - 2b(2a) = 0

⇒ (2a)² + b² + c² - 2c(2a) - 2b(2a) = 0

We can also write (2a)² as (-2a)² because (-) (-) = (+)

⇒ (-2a)² + b² + c² - 2c(2a) - 2b(2a) = 0

We know that (-2a + b + c)² = (-2a)² + b² + c² - 2c(2a) - 2b(2a)

⇒ (-2a + b + c)² = 0

⇒ - 2a + b + c = 0

⇒ b + c = 2a

Hence, proved.

Answer 2

Question :---

• If the roots of the quadratic equation

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

prove that 2a = b + c.

concept used :-----

If A•x^2 + B•x + C = 0 ,is any quadratic equation,

then its discriminant is given by;

D = B^2 - 4•A•C

• If D = 0 , then the given quadratic equation has real and equal roots.

• If D > 0 , then the given quadratic equation has real and distinct roots.

• If D < 0 , then the given quadratic equation has unreal (imaginary) roots...

____________________________

Solution :-----

From Equation (a - b)x² + (b-c)x + (c - a) = 0 ,

we have ,

→ A = (a-b)

→ B = (b-c)

→ C = (c-a)

Since, Roots are Equal, our discriminant will be Equal to Zero ....

→ B² - 4*AC = 0 ,

Putting Values now,

☛ (b-c)² - 4×(a-b)(c-a) = 0

[ we know (x-y)² = x² + y² - 2xy) ]

☛ b² + c² -2bc -4ac + 4a² + 4bc - 4ab = 0

☛ b² + c² +4a² -2bc + 4bc -4ac -4ab = 0

☛b² + c² + 4a² + 2bc - 4ac - 4ab = 0

☛ b² + c² + (-2a)² + 2bc + 2b(-2a) + 2c(-2a) = 0

[ comparing this with the formula

(a+b-c)² = a² + b² + c² + 2ab - 2bc - 2ca we get, ]

☛ (b + c - 2a)²= b² + c² + (-2a)² + 2bc + 2b(-2a) + 2c(-2a)= 0

So, we can say that,

☛ (b +c - 2a) = 0

or ,

☛ b + c = 2a [ Hence, Proved ] .

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★Extra Brainly knowledge :----- ★

[1] ( a + b )² = a² + 2ab + b²

[2] ( a – b )² = a² – 2ab + b²

[3] ( a + b )³ = a³ + 3a² b + 3ab² + b³

[4] ( a – b )³ = a ³ – 3a² b + 3ab² – b³

[5] ( a + b )( a ² - ab + b² ) = a³ + b³

[6] ( a – b )( a ² + ab + b² ) = a³ – b³

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