Question

if the roots of the equation (a-b)x^2 + (b-c)x + (c-a) =0 are equal , then find the value of b+c in terms of a
class 10

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Answer 1

Answer:

if roots of equation (a-b)x2+(b-c)x+(c-a)=0 are equal , prove that b+c = 2a.

Using Discriminant,

D = B2 -4AC as compared with the general quadratic equation Ax2+Bx+C=0

so, A = a-b

B = b-c

C = c-a

For roots to be equal, D=0

(b-c)2 - 4(a-b)(c-a) =0

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

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

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

(2a-b-c)2=0

i.e. 2a-b-c =0

2a= b+c

Answer 2

Given that :

If the roots of the equation $\sf (a-b)x^{2} +(b-c)x +(c-a) = 0$ are equal.

Basic concepts of 10th , related to the nature of root is mentioned below ;

What question says :

Here , We have to have to find the value of b+c in terms of a class 10.

Let's solve it :

By comparing it with the general form of quadratic equation which is ,

⠀⠀⠀⠀⠀$\underline{\bf{px^{2} + qx + r = 0}}$

• p = ( a - b )

• q = ( b - c )

• r = ( c - a )

Then,

Substituting the values , we get :

$\dashrightarrow\:\:\sf {(a - b)x^{2} + ( b - c ) x + (c - a)r = 0}$

D = $\sf q^{2} - 4 pr$ = 0 { Real and equal roots }

$\dashrightarrow\:\:\sf ( b - c )^{2} - 4 \times (a - b) (c - a) = 0$

Using identity :-

$\bullet \: \: {\boxed {\sf (a - b)^{2} = a^{2} + b^{2} - 2ab}}$

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

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

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

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

Taking 2 common ,

$\dashrightarrow\:\:\sf 2 ( 2 a - b - c )^{2} = 0$

We transposed , square on other side :-

$\dashrightarrow\:\:\sf 2a - b - c = 0$

$\dashrightarrow\:\:\sf 2a = b + c$

Hence,

The value of ( b + c ) is ,

$\boxed{\therefore \sf{ (b + c) = 2a }}$

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Class 10th Basic concept :-

Nature of the root :-

Let us we consider the quadratic equation be -

$:\implies\sf ax^{2} + bx + c = 0$ { Where , a ≠ 0 }

Then ,

$\sf \alpha \: and \: \beta$ are the zeroes of the equation :

We know , quadratic formula which is ,

$:\implies\sf \alpha = \dfrac{- b + \sqrt D}{2a}$

And

$:\implies\sf \beta = \dfrac{- b - \sqrt D}{2a}$

The value of the D is = $\sf b^{2} - 4 ac$ ≥ 0.

If D = $\sf b^{2} - 4 ac$ > 0 , then $\sf \alpha \: and \: \beta$ are real { having real roots }.

So, If

$:\implies \sf \alpha - \beta = ( \dfrac{- b - \sqrt D}{2a}) - ( \dfrac{- b + \sqrt D}{2a} )$

$:\implies\sf \dfrac{ - b^{2} + \sqrt D + b + \sqrt D}{ 2a } = \dfrac{2 \sqrt D}{2a} = \dfrac{ \sqrt D}{a}$

→ α - β ≠ 0 → α ≠ β

Therefore,

$\bullet \: \:$ if D = $\sf b^{2} - 4 ac$ > 0 i.e, the discriminant of the equation is positive, then equation has real and distinct roots α and β given as :-

$:\implies \sf \alpha = \dfrac{- b + \sqrt D}{2a}$

And

$:\implies \sf \beta = \dfrac{- b - \sqrt D}{2a}$

$\bullet \: \:$ If D = $\sf b^{2} - 4 ac$ = 0 i.e, the discriminant of equation is zero, then equation has real and equal roots both equal to $\sf - \dfrac{b^{2}}{2a }$

Now, Last but not least , what's the nature of equation when it's discriminant is negative?

$\bullet \: \:$ In this condition, no real root satisfy equation and therefore, their is no real root of the given quadratic equation.

D = $\sf b^{2} - 4 ac$ < 0

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*Note : In the above question , we used 2nd condition i.e, of equal roots.