Question

if the roots of the equation (a-b)x sq +(b-c)x + (c-a)=0 prove tht b+c =2a​

Answer 1

Step-by-step explanation:

We know that,

If the quadratic equation ax²+bx+c=0

whose roots are equal then it's

deteminant is equal to zero.

(a-b)x²+(b-c)x+(c-a)=0

Deteminant =0

(b-c)² -4(a-b)(c-a)==0

b²+c²-2bc-4ac+4a²+4bc-4ab=0

b²+c²+4a²+4bc-4ac-4ab=0

b²+c²+(-2a)²+2bc+2c(-2a)+2(-2a)b=0

(b+c-2a)²=0

b+c-2a=0

Therefore,

b+c=2a

Hence proved.

Answer 2

Look at my final answer!

To find : use discriminant $D=b^2-4ac$

Solution :

Let $a=(a-b)$

And $b=(b-c)$

And then $c=(c-a)$

Use discriminant $D=0$ since roots are equal.

$D=(b-c)^2-4(a-b)(c-a)$

$D=(b-c)^2+4(a-b)(a-c)$

$D=(b^2-2bc+c^2)+4(a^2-ab-ac+bc)$

$D=4a^2+b^2+c^2-4ab+(-2+4)bc-4ac$

$D=4a^2+b^2+c^2-4ab+2bc-4ac$

Stuck? Here's solution :

Identity $x^2+y^2+z^2+2xy+2yz+2zx=(x+y+z)^2$

$(2a)^2+b^2+c^2-2*2a*b+2*b*c-2*2a*c=(2a-b-c)^2$

Now we will use $D=0$

$(2a-b-c)^2=0$

∴$2a-b-c=0$ And further, $b+c=2a$.