Question

If the roots of the equation (a – b) x2 + (b – c) x + (c – a) = 0 are equal, then

find the value of b + c in terms of ‘a’.​

Answer 1

Step-by-step explanation:

This is required solution to the question.

Answer 2

ǫᴜᴇsᴛɪᴏɴ :-

• ★ If the roots of the equation (a – b) x² + (b – c) x + (c – a) = 0 are equal, then find the value of b + c in terms of ‘a’.

✏️ ᴀɴsᴡᴇʀ :-

• ★ b + c = 2a

✏️ sᴛᴇᴘ-ʙʏ-sᴛᴇᴘ ᴇxᴘʟᴀɴᴀᴛɪᴏɴ :-

★ ɢɪᴠᴇɴ :-

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

★ ᴛᴏ ᴘʀᴏᴠɪᴅᴇ :-

• b + c in terms of a

★ Concept Used :-

Let us consider a quadratic equation ax² + bx + c = 0, then roots of the equation are equal if and only if

• Discriminant = 0

It implies

• b² - 4ac = 0.

★ ʀᴀᴛɪᴏɴᴀʟᴇ :-

Given Quadratic equation is

• (a – b) x² + (b – c) x + (c – a) = 0

On comparing with Ax² + Bx + C = 0,

we get

• A = a - b

• B = b - c

• C = c - a

Since,

It is given that,

• Quadratic equation has equal roots,

So,

$\longmapsto\tt \: discriminant \: = \: 0$

$\longmapsto\tt \: {B }^{2} - 4AC = 0$

On substituting the values of A, B and C, we get

$\longmapsto\tt \: {(b - c)}^{2} - 4(a - b)(c - a) = 0$

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

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

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

$\tt \: {b}^{2} + {c}^{2} + {(2a)}^{2} + 2(b)(c) - 2(c)(2a) - 2(2a)(b) = 0$

$\tt \: {b}^{2} + {c}^{2} + {( - 2a)}^{2} + 2(b)(c) - 2(c)(2a) - 2(2a)(b) = 0$

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

$\rm :\implies\:b + c - 2a = 0$

$\bf\implies \:\large \underline{\tt \: \red{ b + c = 2a }}$

Additional Information :-

Let us consider a quadratic equation ax² + bx + c = 0, then nature of roots of the equation depends on Discriminant.

• If Discriminant, D > 0, equation has real and unequal roots.

• If Discriminant, D < 0, equation has no real roots or have complex roots or imaginary roots

• If Discriminant, D = 0, equation has real and equal roots.

• If Discriminant, D is positive and perfect square, equation has real and unequal rational roots.

• If Discriminant, D is positive and not a perfect square, equation has real and irrational roots.