Question

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

Answer 1

Step-by-step explanation:

Given:-

• A quadratic equation (a - b)x² + (b - c)x + (c - a) = 0

• The roots are equal.

To Prove:-

• 2a = b + c

Solution:-

For a quadratic equation ax² + bx + c = 0

If roots are equation

b² - 4ac = 0

Comparing ax² + bx + c = 0 with (a - b)x² + (b - c)x + (c - a) = 0

Here:-

• a = (a - b)

• b = (b - c)

• c = (c - a)

$\sf \longrightarrow b^2 - 4ac = 0$

$\sf \longrightarrow (b - c)^2 - 4 \times (a - b) \times (c - a)= 0$

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

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

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

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

In is the form :- (2a - b - c)²

$\sf \longrightarrow (2a - b - c)^2= 0$

$\sf \longrightarrow 2a - b - c = 0$

$\sf \longrightarrow 2a = b + c$

$\boxed{\bf \leadsto 2a = b + c}$

Hence Proved

Answer 2

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Step-by-step explanation:

Given:-

• A quadratic equation (a - b)x² + (b - c)x + (c - a) = 0

• The roots are equal.

To Prove:-

• 2a = b + c

Solution:-

For a quadratic equation ax² + bx + c= 0

If roots are equation

• b² - 4ac = 0
• Comparing ax² + bx + c = 0 with (a - b)x² + (b - c)x + (c - a) = 0

Here:-

• a = (a - b)

• b = (b - c)

• c = (c - a)

⟶${b}^{2} −4ac=0$

⟶${(b - c)}^{2} −4×(a−b)×(c−a)=0$

⟶${b}^{2} + {c}^{2} −2bc−4(ac−{a^{2}} −bc+ab)=0$

⟶${b}^{2} + {c}^{2} −2bc−4ac+ {4a}^{2} +4bc−4ab=0$

⟶${4a}^{2} + {b}^{2} + {c}^{2} −4ac−2bc+4bc−4ab=0$

⟶${4a}^{2} + {b}^{2} + {c}^{2} −4ac+2bc−4ab=0$

In is the form :- (2a - b - c)²

⟶${ \:(2a−b−c)}^{2} =0$

⟶$2a−b−c=0$

⟶$2a=b+c$

⇝2a = b + c

Hence Proved