Question

if the roots of the equation (b-c)x*2+(c-a)x+(a-b)=0are equal then.Prove that 2b=a+c

Answer 1

if the roots of the equation (b-c)x*2+(c-a)x+(a-b)=0are equal then.Prove that 2b=a+c


your answer is here...

(b-c)x²+(c-a)x+(a-b)=0

Comparing with quadratic equation

Ax²+Bx+C=0

A=(b-c),
B=c-a,
C=a-b

Discriminate when roots are equal

D=B²-4AC=0

D=(c-a)²−4(b-c)(a-b)=0

D=(c²+a²−2ac)-4(ba-ac-b²+bc)=0

D=c²+a²−2ac-4ab+4ac+4b²-4bc=0

c²+a²+2ac-4b(a+c)+4b²=0

(a+c)²-4b(a+c)+4b²=0

[(a+c)-2b]²=0

a+c=2b

Answer 2

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

Given:

• A quadratic equation
• $(b - c) {x}^{2} + (c - a)x + (a - b) = 0$

To find:

• Prove that prove that 2b=a+c, if roots of the equation are equal.

Solution:

Concept to be used:

If roots of quadratic equation $\bf e {x}^{2} + fx + g = 0$ are equal than its discriminate $\bf D = {f}^{2} - 4eg$ is zero.

Step 1:

Write coefficients of x², x and constant term.

$e = b - c \\ f = c - a \\ g = a - b$

Step 2:

Put D=0

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

or

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

or

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

it is resembles to the identity

$\bf (x + y + z)^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2xz$

so,

${a}^{2} +( { - 2b)}^{2} + {c}^{2} + 2a( - 2b) + 2( - 2b)c + 2ac= 0$

thus,

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

taking square root both sides.

$a - 2b + c = 0$

or

$a + c = 2b$

or

$\bf 2b = a + c$

Thus,

It has been proven that 2b=a+c.

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