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

GIVEN :–

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

TO PROVE :–

• 2a = b + c

SOLUTION :–

• If a quadratic equation ax² + bx + c = 0 & it's roots are equal then Discriminant of quadratic equation will be zero.

$\\ \implies\large { \boxed{ \bf D = 0}}$

$\\ \implies\large { \boxed{ \bf {b}^{2} - 4ac = 0}}$

• Now put the values –

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

$\\ \implies\bf {(b - c)}^{2} = 4(a - b)(c - a)$

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

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

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

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

• We should write this as –

$\\ \implies\bf {(2a - b - c)}^{2} = 0$

$\\ \implies\bf 2a - b - c= 0$

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

$\\ \large\longrightarrow{ \boxed{\bf Hence \: \:Proved }}$

Answer 2

➠GIVEN:-

The roots of the quadratic equation (a-b)x2+(b-c)x+(c-α) = 0 are equal.

➠TO PROVE:-

2a=b+c

➠PROOF:-

•Let the given equation is in the form of Ax²+Bx+C

Where ,

•A=(a-b),

•B=(b-c) and

• C=(c-a)

Apply below formula for finding Discriminant of the quadratic equation

•D=B²-4AC

∴D=(b-c)²-4(a-b)(c-a)

For equal roots we must have Discriminant, D=0

Now ,D=0

⇛(b-c)²-4(a-b)(c-a)=0

⇛4a²+b²+c²-4ab+2bc-4ac=0

⇛(-2a)²+b²+c²+2(-2a)b+2bc+2c(-2a)=0

⇛(-2a+b+c)²=0

⇛-2a+b+c=0

⇛b+c=2a

Or 2a=b+c

Hence,proved